{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0001","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0106e25774aaee750b53ba3a466c8a340571ef3382ccccf9c14e53eb9cdb60cc","family_id":"qf-mod34-series-partial-sum","seed":202611033001,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,2],[-1,3]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/2, -1/3]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/2, -1/3], 1/6, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/2, -1/3], 1/6, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0002","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3db1a91af951a04759d19411366dc890e98bef2b4385b1099389b7a74c3d3c2b","family_id":"qf-mod34-series-partial-sum","seed":202611033002,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,3],[-1,4]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/3, -1/4]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/3, -1/4], 1/12, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/3, -1/4], 1/12, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0003","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2380b600ad1d2061a5d1963f750d4532e502d90cfb79c81f9aa55ce1250313e","family_id":"qf-mod34-series-partial-sum","seed":202611033003,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,4],[-1,5]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/4, -1/5]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/4, -1/5], 1/20, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/4, -1/5], 1/20, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0004","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4d4c862fe2fa43b6f79555d17ff5a47d5a204eb86daac457641b9b7e933c79c","family_id":"qf-mod34-series-partial-sum","seed":202611033004,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,5],[-1,6]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/5, -1/6]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/5, -1/6], 1/30, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/5, -1/6], 1/30, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0005","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01a0c7d6e3c67e8b152321bf011782e8445f327bd085c38a53282e7efaf9d28c","family_id":"qf-mod34-series-partial-sum","seed":202611033005,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,6],[-1,7]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/6, -1/7]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/6, -1/7], 1/42, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/6, -1/7], 1/42, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0006","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"243d272cc4d3dcef8a0db965f6d48eb950d5aa3f63d2079dbac61d8ee0b05e0c","family_id":"qf-mod34-series-partial-sum","seed":202611033006,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,7],[-1,8]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/7, -1/8]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/7, -1/8], 1/56, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/7, -1/8], 1/56, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0007","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3ac7bbe800eff575003287b431a11c87220f53838dd9276b7853a276bda299f","family_id":"qf-mod34-series-partial-sum","seed":202611033007,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,8],[-1,9]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/8, -1/9]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/8, -1/9], 1/72, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/8, -1/9], 1/72, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0008","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b53582c58f99511805433ae3e87c8ba0725b56d0c526c990157a10191bd18116","family_id":"qf-mod34-series-partial-sum","seed":202611033008,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,9],[-1,10]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/9, -1/10]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/9, -1/10], 1/90, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/9, -1/10], 1/90, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0009","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efd61f615591695242ec7eb67e0dd87ce116dcc9334b989c42daebc415bc4085","family_id":"qf-mod34-series-partial-sum","seed":202611033009,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,10],[-1,11]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/10, -1/11]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/10, -1/11], 1/110, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/10, -1/11], 1/110, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0010","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54d1166229b260e8e4e30732d2d15797433a7719219bc0f103d4dcb6c9460124","family_id":"qf-mod34-series-partial-sum","seed":202611033010,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,11],[-1,12]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/11, -1/12]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/11, -1/12], 1/132, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/11, -1/12], 1/132, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0011","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"775f46f3523162e3ba3d7d265c53ab6c9c3e3993727baedceb7fcd5b0cd32dd9","family_id":"qf-mod34-series-partial-sum","seed":202611033011,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,12],[-1,13]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/12, -1/13]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/12, -1/13], 1/156, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/12, -1/13], 1/156, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0012","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60545036247ee15954e7adfbe53dc2156f03d0080cb011e7f69c76c788f5dd0b","family_id":"qf-mod34-series-partial-sum","seed":202611033012,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,13],[-1,14]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/13, -1/14]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/13, -1/14], 1/182, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/13, -1/14], 1/182, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0013","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b80b3d94b8d589f9ad039250f9267931a0dff1511a4e521d79815d785f22dc24","family_id":"qf-mod34-series-partial-sum","seed":202611033013,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,14],[-1,15]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/14, -1/15]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/14, -1/15], 1/210, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/14, -1/15], 1/210, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0014","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"198157b4a4876b8aafac73785b9b063ed63c026401590554cda6166013afb34c","family_id":"qf-mod34-series-partial-sum","seed":202611033014,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,15],[-1,16]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/15, -1/16]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/15, -1/16], 1/240, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/15, -1/16], 1/240, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0015","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83f25e1853ccfe8b734ee8acdb54f2b61d65745f806a3f1c9c2dd017826c631e","family_id":"qf-mod34-series-partial-sum","seed":202611033015,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,16],[-1,17]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/16, -1/17]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/16, -1/17], 1/272, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/16, -1/17], 1/272, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0016","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2109e9688f598d2ddb8b4e72271021245e3779f1af45281620dcc69c8bc06af","family_id":"qf-mod34-series-partial-sum","seed":202611033016,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,17],[-1,18]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/17, -1/18]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/17, -1/18], 1/306, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/17, -1/18], 1/306, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0017","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c98b8555093c4d0af9a4a909b671644b12394c1abf600ab56dc76ab8c98bc1f8","family_id":"qf-mod34-series-partial-sum","seed":202611033017,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,18],[-1,19]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/18, -1/19]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/18, -1/19], 1/342, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/18, -1/19], 1/342, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0018","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9acfb237adfef3e300cdab94e702d9cc4b9bff8c8888da8bcc00a988bce9466f","family_id":"qf-mod34-series-partial-sum","seed":202611033018,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,19],[-1,20]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/19, -1/20]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/19, -1/20], 1/380, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/19, -1/20], 1/380, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0019","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3358b2aebed9291b5cebbbd102c562dff02dc4a2d3c60dffe8eba96eafd309b","family_id":"qf-mod34-series-partial-sum","seed":202611033019,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,20],[-1,21]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/20, -1/21]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/20, -1/21], 1/420, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/20, -1/21], 1/420, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0020","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91d217a12280c1fedee1d80689d120a35475aac216ce0f92abb2cb422afca283","family_id":"qf-mod34-series-partial-sum","seed":202611033020,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,21],[-1,22]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/21, -1/22]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/21, -1/22], 1/462, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/21, -1/22], 1/462, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0021","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dfecaa9f3d59d3f99c3f9db18110ca0a15650d94ff5b3c3f12f0e525c739ab72","family_id":"qf-mod34-series-partial-sum","seed":202611033021,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,22],[-1,23]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/22, -1/23]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/22, -1/23], 1/506, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/22, -1/23], 1/506, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0022","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68204f1a141a7c597334cc1727f3c30f0add0c1ddced7dbd3f4152b3418672a2","family_id":"qf-mod34-series-partial-sum","seed":202611033022,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,23],[-1,24]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/23, -1/24]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/23, -1/24], 1/552, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/23, -1/24], 1/552, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0023","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c88d5a73c6b3a19f08566a6d93d8bbc8a476a0c1ea1a634f96ca52e47a39dc2","family_id":"qf-mod34-series-partial-sum","seed":202611033023,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,24],[-1,25]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/24, -1/25]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/24, -1/25], 1/600, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/24, -1/25], 1/600, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0024","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb647629c652680e5041e8bb7a74a9776ebb7dcf968e8674f0eec15d495b1295","family_id":"qf-mod34-series-partial-sum","seed":202611033024,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,25],[-1,26]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/25, -1/26]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/25, -1/26], 1/650, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/25, -1/26], 1/650, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0025","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce8ac6ace532970b8b89a57fc79745a90f70703eeb085db564f3c6543c377310","family_id":"qf-mod34-series-partial-sum","seed":202611033025,"semantic_template":"qf-mod34-series-partial-sum/{\"terms\":[[1,26],[-1,27]]}"},"fields":{"title":"A Finite Partial Sum Is Not an Infinite-Series Limit","stem":"Use the ordered explicit rational terms=[1/26, -1/27]. Give (terms,partial_sum,count,finite_partial_sum_only), reducing the supplied term values and computed finite sum exactly.","answer":{"kind":"expression","value":"([1/26, -1/27], 1/702, 2, finite_partial_sum_only)","explanation":"Add only these listed terms with exact rational arithmetic. An empty list has sum zero. This finite operation supplies no conclusion about an unspecified continuation of the series. The exact result, retaining the requested order, is ([1/26, -1/27], 1/702, 2, finite_partial_sum_only)."},"skills":["A Finite Partial Sum Is Not an Infinite-Series Limit"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0026","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f818fcf43a6c6a7433c117678696c2b12a4396ca5d6ee5b3a83e36773e7cec3","family_id":"qf-mod34-series-geometric-sum","seed":202611033026,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[1,2],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=1/2 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(1/2, 1/2, 1, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (1/2, 1/2, 1, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0027","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a6d3cf2672a0d72392f6f9cb445147473a622fd427fe1f226e9a09d77073b90","family_id":"qf-mod34-series-geometric-sum","seed":202611033027,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[2,3],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=2/3 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(2/3, 1/2, 4/3, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (2/3, 1/2, 4/3, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0028","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f289c14981f3fe2bdde61835857240035c94a8140d06f7500151e58e2741fee","family_id":"qf-mod34-series-geometric-sum","seed":202611033028,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[3,4],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=3/4 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(3/4, 1/2, 3/2, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (3/4, 1/2, 3/2, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0029","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8ac244bc3a69793e025ad8ef907f7b3a58695adb283227bcf5da051fcbaffbc","family_id":"qf-mod34-series-geometric-sum","seed":202611033029,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[4,5],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=4/5 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(4/5, 1/2, 8/5, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (4/5, 1/2, 8/5, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0030","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae008eddeaca924b1110d66b1f7a7b22a4ce4383cf17e378bf899afddb2f245a","family_id":"qf-mod34-series-geometric-sum","seed":202611033030,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[5,6],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=5/6 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(5/6, 1/2, 5/3, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (5/6, 1/2, 5/3, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0031","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc9d9f2470fa75f576848c23321ec6d98bc6d5fc2b17b3541a18043d40c3419a","family_id":"qf-mod34-series-geometric-sum","seed":202611033031,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[6,7],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=6/7 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(6/7, 1/2, 12/7, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (6/7, 1/2, 12/7, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0032","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0bec2835a7e18d5e1b1b534c896b0b30ab14d8387bdcf2e6837daf284e32d285","family_id":"qf-mod34-series-geometric-sum","seed":202611033032,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[7,8],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=7/8 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(7/8, 1/2, 7/4, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (7/8, 1/2, 7/4, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0033","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cc13b341ec35927ce1b0c19dbee727d62705787c55ae4b107f953dc823e1818","family_id":"qf-mod34-series-geometric-sum","seed":202611033033,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[8,9],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=8/9 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(8/9, 1/2, 16/9, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (8/9, 1/2, 16/9, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0034","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e95f6e208d8b93d7b44bf7b89dd3bce7c2074f80e083028faaefe9cb64443d00","family_id":"qf-mod34-series-geometric-sum","seed":202611033034,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[9,10],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=9/10 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(9/10, 1/2, 9/5, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (9/10, 1/2, 9/5, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0035","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60c8b63ac30a305d52efd2875921e2247a0596b00aade02fc47f7aa30e57dbce","family_id":"qf-mod34-series-geometric-sum","seed":202611033035,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[10,11],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=10/11 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(10/11, 1/2, 20/11, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (10/11, 1/2, 20/11, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0036","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af2ce567bec91545cfcf556a5a9a81ada978cf052d5b830151ea2d7e055843a1","family_id":"qf-mod34-series-geometric-sum","seed":202611033036,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[11,12],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=11/12 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(11/12, 1/2, 11/6, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (11/12, 1/2, 11/6, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0037","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04d94f7b0dd26e2d106448c24dd264820735c5a6eed3e220c1db5e75102f3982","family_id":"qf-mod34-series-geometric-sum","seed":202611033037,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[12,13],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=12/13 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(12/13, 1/2, 24/13, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (12/13, 1/2, 24/13, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0038","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"316fa80990c003ebb86115281b75b7bee186a43fb83ebbc8359ff54270208dd5","family_id":"qf-mod34-series-geometric-sum","seed":202611033038,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[13,14],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=13/14 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(13/14, 1/2, 13/7, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (13/14, 1/2, 13/7, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0039","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2941ec49c5e41169663aa400c6c151cd199b78a18daad915b20cf84b727509e","family_id":"qf-mod34-series-geometric-sum","seed":202611033039,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[14,15],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=14/15 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(14/15, 1/2, 28/15, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (14/15, 1/2, 28/15, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0040","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"992c0a6c634db1e7111fd6a94b87408f6651940c28f193dd8c6bd229752174db","family_id":"qf-mod34-series-geometric-sum","seed":202611033040,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[15,16],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=15/16 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(15/16, 1/2, 15/8, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (15/16, 1/2, 15/8, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0041","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e37d08291cab1dd141d74be8d5c925d871411efdaa1d943a54721b3be2ffa0aa","family_id":"qf-mod34-series-geometric-sum","seed":202611033041,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[16,17],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=16/17 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(16/17, 1/2, 32/17, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (16/17, 1/2, 32/17, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0042","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3bdf5ac3f5361eb75491f6f377a719adf9d79d2f6860c24c7f54f1d3bcc81746","family_id":"qf-mod34-series-geometric-sum","seed":202611033042,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[17,18],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=17/18 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(17/18, 1/2, 17/9, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (17/18, 1/2, 17/9, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0043","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1fd826652741fe090731b0b23918f839c6cb146ca264b7f00d77322ba8940c5","family_id":"qf-mod34-series-geometric-sum","seed":202611033043,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[18,19],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=18/19 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(18/19, 1/2, 36/19, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (18/19, 1/2, 36/19, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0044","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94d2fec513c8a18270da4b0d4305af2dfc6684e40b03a4a463b2c59529733e75","family_id":"qf-mod34-series-geometric-sum","seed":202611033044,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[19,20],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=19/20 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(19/20, 1/2, 19/10, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (19/20, 1/2, 19/10, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0045","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a404aa167c1d3ccb06bb9f2a7cbac487bcb62ea571f9aea21eccbfea16c74d34","family_id":"qf-mod34-series-geometric-sum","seed":202611033045,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[20,21],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=20/21 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(20/21, 1/2, 40/21, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (20/21, 1/2, 40/21, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0046","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd609c17658ffeb5b5bccb58fafe240c121ac1eedd78695cc7778c538cf0e91c","family_id":"qf-mod34-series-geometric-sum","seed":202611033046,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[21,22],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=21/22 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(21/22, 1/2, 21/11, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (21/22, 1/2, 21/11, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0047","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6b140d44efbe1cf37623eb001c8eacccffca5610001c55a0bd3f3a1234ca77d","family_id":"qf-mod34-series-geometric-sum","seed":202611033047,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[22,23],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=22/23 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(22/23, 1/2, 44/23, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (22/23, 1/2, 44/23, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0048","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1642e0230d9b47524ff99180a8a4a6d13ca80e836e05f57f68125687b75ff2f9","family_id":"qf-mod34-series-geometric-sum","seed":202611033048,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[23,24],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=23/24 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(23/24, 1/2, 23/12, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (23/24, 1/2, 23/12, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0049","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d3962df1bfe28e27eb762f4cd6e2445567e1d972c8575e1d47daacd1da247c0","family_id":"qf-mod34-series-geometric-sum","seed":202611033049,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[24,25],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=24/25 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(24/25, 1/2, 48/25, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (24/25, 1/2, 48/25, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0050","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80160e878b0872e4e62a319593adf427c6758e77b2f0cd9664e6cb9a4368e098","family_id":"qf-mod34-series-geometric-sum","seed":202611033050,"semantic_template":"qf-mod34-series-geometric-sum/{\"a\":[25,26],\"r\":[1,2]}"},"fields":{"title":"The Geometric-Series Formula Retains Its Absolute-Ratio Premise","stem":"Use the infinite series with term a*r^k for k=0,1,..., exact a=25/26 and r=1/2, requiring |r|<1. Give (a,r,infinite_sum,absolute_ratio_below_one_symbolic_tail).","answer":{"kind":"expression","value":"(25/26, 1/2, 25/13, absolute_ratio_below_one_symbolic_tail)","explanation":"Under the stated absolute-ratio premise calculate a/(1−r) exactly and use the explicit geometric tail certificate. Inputs outside this certificate domain are rejected; passing finitely many partial sums is not the proof. The exact result, retaining the requested order, is (25/26, 1/2, 25/13, absolute_ratio_below_one_symbolic_tail)."},"skills":["The Geometric-Series Formula Retains Its Absolute-Ratio Premise"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0051","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfee8d0caae66a8f2e4d74b45bce610af6c40003ee8ecd9c618301f7795ae2d3","family_id":"qf-mod34-series-telescoping-sum","seed":202611033051,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":1,\"n\":1,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=1, positive shift s=2, and n=1 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(1, 2, 1, 1/6, 1/3, 1/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (1, 2, 1, 1/6, 1/3, 1/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0052","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5233ac49efd00313325f1177a2f70ce08e336dd446353a66662358beb908962d","family_id":"qf-mod34-series-telescoping-sum","seed":202611033052,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":2,\"n\":2,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=2, positive shift s=2, and n=2 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(2, 2, 2, 1/2, 2/4, 2/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (2, 2, 2, 1/2, 2/4, 2/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0053","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"646226343a3328a5cf4d9b277b5e248b5130ec5f1c080928ba3a173c8466de71","family_id":"qf-mod34-series-telescoping-sum","seed":202611033053,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":3,\"n\":3,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=3, positive shift s=2, and n=3 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(3, 2, 3, 9/10, 3/5, 3/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (3, 2, 3, 9/10, 3/5, 3/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0054","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9c597a72091270dbccd255b939b84bbc9724e6cb4079f4b67c364eec3ef58a4","family_id":"qf-mod34-series-telescoping-sum","seed":202611033054,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":4,\"n\":4,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=4, positive shift s=2, and n=4 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(4, 2, 4, 4/3, 4/6, 4/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (4, 2, 4, 4/3, 4/6, 4/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0055","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f8b6bb3ee837d65771c33e94b9e0ba43e6db6a4894f7b9fecada375549e1794","family_id":"qf-mod34-series-telescoping-sum","seed":202611033055,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":5,\"n\":5,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=5, positive shift s=2, and n=5 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(5, 2, 5, 25/14, 5/7, 5/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (5, 2, 5, 25/14, 5/7, 5/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0056","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd35d00a4a85e025d701bfdef75f48107bb09e9cab8335691d4a0feb6adc63bd","family_id":"qf-mod34-series-telescoping-sum","seed":202611033056,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":6,\"n\":6,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=6, positive shift s=2, and n=6 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(6, 2, 6, 9/4, 6/8, 6/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (6, 2, 6, 9/4, 6/8, 6/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0057","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"650ab70f4ee80a169e2737795e5785dc0951cd501bf4cd724b50a112375c25ed","family_id":"qf-mod34-series-telescoping-sum","seed":202611033057,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":7,\"n\":7,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=7, positive shift s=2, and n=7 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(7, 2, 7, 49/18, 7/9, 7/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (7, 2, 7, 49/18, 7/9, 7/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0058","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e35e6f132bd35a1798ba4095ae6b45870adfc32af95a1e013da6bbf14772e7e","family_id":"qf-mod34-series-telescoping-sum","seed":202611033058,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":8,\"n\":8,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=8, positive shift s=2, and n=8 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(8, 2, 8, 16/5, 8/10, 8/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (8, 2, 8, 16/5, 8/10, 8/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0059","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"608eeb9ffc4a09fe4835187ef1f3db5177c4b9a6bad215878e048af3be44e0b1","family_id":"qf-mod34-series-telescoping-sum","seed":202611033059,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":9,\"n\":9,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=9, positive shift s=2, and n=9 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(9, 2, 9, 81/22, 9/11, 9/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (9, 2, 9, 81/22, 9/11, 9/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0060","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0aafa5f84c3966fdc5ffb48406f77112dbbe87f1a4e374818d445cf70a330d86","family_id":"qf-mod34-series-telescoping-sum","seed":202611033060,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":10,\"n\":10,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=10, positive shift s=2, and n=10 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(10, 2, 10, 25/6, 10/12, 10/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (10, 2, 10, 25/6, 10/12, 10/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0061","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef11622375e8a06846ef8aaaa6cd1c204de316fc7fdd8dd2cccd73964552ebc0","family_id":"qf-mod34-series-telescoping-sum","seed":202611033061,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":11,\"n\":11,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=11, positive shift s=2, and n=11 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(11, 2, 11, 121/26, 11/13, 11/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (11, 2, 11, 121/26, 11/13, 11/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0062","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9607681fce56c9a3757a5c9d0e6c756274dd0385bca2a6fe9702905d5f1d3e2","family_id":"qf-mod34-series-telescoping-sum","seed":202611033062,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":12,\"n\":12,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=12, positive shift s=2, and n=12 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(12, 2, 12, 36/7, 12/14, 12/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (12, 2, 12, 36/7, 12/14, 12/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0063","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d327215aec9281a684b8ca4c64f80c3422255533d99fd594f0036a6fb156d9f2","family_id":"qf-mod34-series-telescoping-sum","seed":202611033063,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":13,\"n\":13,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=13, positive shift s=2, and n=13 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(13, 2, 13, 169/30, 13/15, 13/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (13, 2, 13, 169/30, 13/15, 13/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0064","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c5b26d98a3287e15a7c6d0d8d05a5f8080747eb799810104bc8eef6f5d7b6bc","family_id":"qf-mod34-series-telescoping-sum","seed":202611033064,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":14,\"n\":14,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=14, positive shift s=2, and n=14 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(14, 2, 14, 49/8, 14/16, 14/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (14, 2, 14, 49/8, 14/16, 14/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0065","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb08be13d334c15dbf4d08bda64365bc1b8315bddb7b18e4c0856386f5dbc6ff","family_id":"qf-mod34-series-telescoping-sum","seed":202611033065,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":15,\"n\":15,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=15, positive shift s=2, and n=15 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(15, 2, 15, 225/34, 15/17, 15/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (15, 2, 15, 225/34, 15/17, 15/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0066","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c35f415ca45e3bf1245b3b336fe6c5caeae5fae9fec0810f80955683657098ac","family_id":"qf-mod34-series-telescoping-sum","seed":202611033066,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":16,\"n\":16,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=16, positive shift s=2, and n=16 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(16, 2, 16, 64/9, 16/18, 16/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (16, 2, 16, 64/9, 16/18, 16/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0067","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5987efd6e07ece66398ed375e3657003e4e59a30ee9b4fcd03640179377bbc8","family_id":"qf-mod34-series-telescoping-sum","seed":202611033067,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":17,\"n\":17,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=17, positive shift s=2, and n=17 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(17, 2, 17, 289/38, 17/19, 17/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (17, 2, 17, 289/38, 17/19, 17/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0068","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1dc091f10549be0a4472dda6ee1628713ee28a94e787d5b05c1e844a0cc56e82","family_id":"qf-mod34-series-telescoping-sum","seed":202611033068,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":18,\"n\":18,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=18, positive shift s=2, and n=18 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0069","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6cb4af4a7e22d9b27865507ed7e5aa6804ef46dee3cac195413600adc1bb871e","family_id":"qf-mod34-series-telescoping-sum","seed":202611033069,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":19,\"n\":19,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=19, positive shift s=2, and n=19 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(19, 2, 19, 361/42, 19/21, 19/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (19, 2, 19, 361/42, 19/21, 19/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0070","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75c0460da94429697f8a7a5ad04c0a07227fc68498886c977efee2a6e57a0280","family_id":"qf-mod34-series-telescoping-sum","seed":202611033070,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":20,\"n\":20,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=20, positive shift s=2, and n=20 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(20, 2, 20, 100/11, 20/22, 20/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (20, 2, 20, 100/11, 20/22, 20/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0071","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f06c6cc6e1830ca89b9367caafd06ebd2f9875a15fc61ede494bb6bbadf592cc","family_id":"qf-mod34-series-telescoping-sum","seed":202611033071,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":21,\"n\":21,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=21, positive shift s=2, and n=21 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(21, 2, 21, 441/46, 21/23, 21/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (21, 2, 21, 441/46, 21/23, 21/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0072","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45b6cf1702a8a544c5e3f0e490e35150221bddbbf8eb1fb5f39ac39f0a6be035","family_id":"qf-mod34-series-telescoping-sum","seed":202611033072,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":22,\"n\":22,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=22, positive shift s=2, and n=22 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(22, 2, 22, 121/12, 22/24, 22/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (22, 2, 22, 121/12, 22/24, 22/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0073","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c90880f22d1a329ad011e25ba404bffd65e575ff91716db813b603125988bebe","family_id":"qf-mod34-series-telescoping-sum","seed":202611033073,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":23,\"n\":23,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=23, positive shift s=2, and n=23 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(23, 2, 23, 529/50, 23/25, 23/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (23, 2, 23, 529/50, 23/25, 23/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0074","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2da62b97692a50eefd0d989fc7063e168c9c394a997d4e8b95ea228be770843","family_id":"qf-mod34-series-telescoping-sum","seed":202611033074,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":24,\"n\":24,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=24, positive shift s=2, and n=24 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(24, 2, 24, 144/13, 24/26, 24/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (24, 2, 24, 144/13, 24/26, 24/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0075","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43240658d81cd6f4f4391221cf71cb2d867853553222ff6e2e05f2f957d0489c","family_id":"qf-mod34-series-telescoping-sum","seed":202611033075,"semantic_template":"qf-mod34-series-telescoping-sum/{\"c\":25,\"n\":25,\"shift\":2}"},"fields":{"title":"Retain Both the Telescoping Remainder and the Source Ratio Text","stem":"Use c=25, positive shift s=2, and n=25 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.","answer":{"kind":"expression","value":"(25, 2, 25, 625/54, 25/27, 25/2, exact_telescoping_remainder)","explanation":"Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (25, 2, 25, 625/54, 25/27, 25/2, exact_telescoping_remainder)."},"skills":["Retain Both the Telescoping Remainder and the Source Ratio Text"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0076","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7bce0ac133013ba642d047eb868c7c131807eb742a825ad576988664424083a1","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033076,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":1}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=1 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(1, 5, [1, 1/2, 1/3, 1/4, 1/5], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (1, 5, [1, 1/2, 1/3, 1/4, 1/5], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0077","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f36f4a9518d524520662007b4dc794e2f4f9ad5c1d24d276a29aabbec1266ee","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033077,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":2}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=2 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(2, 5, [1/2, 1/3, 1/4, 1/5, 1/6], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (2, 5, [1/2, 1/3, 1/4, 1/5, 1/6], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0078","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"903c66c39e34ed9148531a6e359997616cd4bf27f8a7325bccfd13376c4acfad","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033078,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":3}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=3 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(3, 5, [1/3, 1/4, 1/5, 1/6, 1/7], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (3, 5, [1/3, 1/4, 1/5, 1/6, 1/7], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0079","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a07bec6d73950af6c503c59ac15d4369050ac3f93eed7f37782b3ff5c65294f0","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033079,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":4}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=4 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(4, 5, [1/4, 1/5, 1/6, 1/7, 1/8], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (4, 5, [1/4, 1/5, 1/6, 1/7, 1/8], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0080","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"278305fc180824a26ca7eec7fd348d03611371cecc37c114ee7ed6b76a344cad","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033080,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":5}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=5 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(5, 5, [1/5, 1/6, 1/7, 1/8, 1/9], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (5, 5, [1/5, 1/6, 1/7, 1/8, 1/9], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0081","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64a231661352c7982fc3b2fc156c051688b86096514920629992f130ea0322ab","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033081,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":6}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=6 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(6, 5, [1/6, 1/7, 1/8, 1/9, 1/10], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (6, 5, [1/6, 1/7, 1/8, 1/9, 1/10], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0082","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5a4452ea7d82ba7773b4f53595146f9358c9152b9f51760ec79349f478d4df5","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033082,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":7}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=7 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(7, 5, [1/7, 1/8, 1/9, 1/10, 1/11], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (7, 5, [1/7, 1/8, 1/9, 1/10, 1/11], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0083","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a2ae6971a39917fe4ed3c14abb68bb51ff9d97ab02b2a6a38fd237e645a896b","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033083,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":8}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=8 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(8, 5, [1/8, 1/9, 1/10, 1/11, 1/12], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (8, 5, [1/8, 1/9, 1/10, 1/11, 1/12], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0084","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b440d93ed680cad749d07d346fa35f77031cbfef08c49b90c889c48923bd4d30","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033084,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":9}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=9 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(9, 5, [1/9, 1/10, 1/11, 1/12, 1/13], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (9, 5, [1/9, 1/10, 1/11, 1/12, 1/13], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0085","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d4b4a823ae96eb6f93b19a7635adfc180739889ed5eaa4d95c6f8a2d9a75be2","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033085,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":10}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=10 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(10, 5, [1/10, 1/11, 1/12, 1/13, 1/14], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (10, 5, [1/10, 1/11, 1/12, 1/13, 1/14], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0086","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a2a449e9070090732210515d93b14c13cbf19e3b1eb8acab32553689289cbb2","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033086,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":11}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=11 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(11, 5, [1/11, 1/12, 1/13, 1/14, 1/15], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (11, 5, [1/11, 1/12, 1/13, 1/14, 1/15], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0087","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00f7ec093a9569d5bf0043000dad7da5c625f52240da1a669ba482d630fcf7df","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033087,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":12}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=12 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(12, 5, [1/12, 1/13, 1/14, 1/15, 1/16], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (12, 5, [1/12, 1/13, 1/14, 1/15, 1/16], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0088","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e137ab09cdb9d8dd821d53e9def993004b2a2430f69d956b181a70720091f96f","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033088,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":13}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=13 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(13, 5, [1/13, 1/14, 1/15, 1/16, 1/17], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (13, 5, [1/13, 1/14, 1/15, 1/16, 1/17], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0089","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"57b268ffe41e685e72b0d7560ae23ab43b88bfcb940ecd3814cdb355f90d2f08","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033089,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":14}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=14 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(14, 5, [1/14, 1/15, 1/16, 1/17, 1/18], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (14, 5, [1/14, 1/15, 1/16, 1/17, 1/18], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0090","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e2026e0ce4d20cf3216df7d44111f92a32234fc3406659cc19af811b0a36d72","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033090,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":15}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=15 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(15, 5, [1/15, 1/16, 1/17, 1/18, 1/19], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (15, 5, [1/15, 1/16, 1/17, 1/18, 1/19], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0091","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef2cb6d996ac1a8efe350e41632613f2002915cc0944a5d103d4d6c9aab4e3b9","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033091,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":16}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=16 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(16, 5, [1/16, 1/17, 1/18, 1/19, 1/20], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (16, 5, [1/16, 1/17, 1/18, 1/19, 1/20], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0092","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a544040955c6c02913d2864823acb8feff01a34606498a0d061cc71000eaeacd","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033092,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":17}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=17 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(17, 5, [1/17, 1/18, 1/19, 1/20, 1/21], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (17, 5, [1/17, 1/18, 1/19, 1/20, 1/21], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0093","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35fdc29350fdb55106fa580fba84fff01bd4e312b2593048241e9314b8a1cd5d","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033093,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":18}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=18 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(18, 5, [1/18, 1/19, 1/20, 1/21, 1/22], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (18, 5, [1/18, 1/19, 1/20, 1/21, 1/22], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0094","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e1d12a837aebc3c82f6bb55d0cdafc557ef0c8ffdffd6b8e77b25ce8a692a90","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033094,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":19}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=19 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(19, 5, [1/19, 1/20, 1/21, 1/22, 1/23], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (19, 5, [1/19, 1/20, 1/21, 1/22, 1/23], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0095","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83b69a547d6fc9354719092a6dc82e92bbd6c6e87fac6b60cf0e282ca736ce21","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033095,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":20}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=20 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(20, 5, [1/20, 1/21, 1/22, 1/23, 1/24], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (20, 5, [1/20, 1/21, 1/22, 1/23, 1/24], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0096","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3d6e6e90acb0c7ca32880f875fdda1298eaf256a385ff7ce1b7154c97b4d35f","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033096,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":21}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=21 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(21, 5, [1/21, 1/22, 1/23, 1/24, 1/25], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (21, 5, [1/21, 1/22, 1/23, 1/24, 1/25], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0097","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae06cb9159284aabdda1a6463d48429e9ba5f3bc6473a066edf8f09f5c49f281","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033097,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":22}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=22 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(22, 5, [1/22, 1/23, 1/24, 1/25, 1/26], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (22, 5, [1/22, 1/23, 1/24, 1/25, 1/26], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0098","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efe54b0c1aee17a34d42afac8c7a45ef402818206d5c11c86e987e4cbf4440f2","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033098,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":23}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=23 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(23, 5, [1/23, 1/24, 1/25, 1/26, 1/27], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (23, 5, [1/23, 1/24, 1/25, 1/26, 1/27], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0099","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c66139364d07daa1f491b9b2eae28598479974ab0dcbc507c4584e2d7ef29cfe","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033099,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":24}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=24 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(24, 5, [1/24, 1/25, 1/26, 1/27, 1/28], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (24, 5, [1/24, 1/25, 1/26, 1/27, 1/28], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261103-r03-infinite-series-0100","batch_id":"batch-20261014-advance-20261103-r03-infinite-series-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"626a3066b1889104249fba4174d6de9f47d97ad3c1ab30e68892a2628dc36686","family_id":"qf-mod34-series-term-zero-boundary","seed":202611033100,"semantic_template":"qf-mod34-series-term-zero-boundary/{\"prefix_length\":5,\"shift\":25}"},"fields":{"title":"Terms Tending to Zero Do Not Make a Harmonic Tail Summable","stem":"Use the shifted harmonic terms 1/(n+shift), n=0,1,..., with positive shift=25 and displayed prefix_length=5. Give (shift,prefix_length,terms,terms_tend_zero?,series_converges?,terms_tend_to_zero_but_shifted_harmonic_series_diverges).","answer":{"kind":"expression","value":"(25, 5, [1/25, 1/26, 1/27, 1/28, 1/29], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)","explanation":"The displayed terms are only a finite prefix, while the named shifted-harmonic argument supplies the divergence conclusion. Terms tend to zero, a necessary but not sufficient condition for series convergence. Do not confuse the limit of terms with the limit of partial sums. The exact result, retaining the requested order, is (25, 5, [1/25, 1/26, 1/27, 1/28, 1/29], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)."},"skills":["Terms Tending to Zero Do Not Make a Harmonic Tail Summable"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}