# 級數比值判別法基礎 · 脈衝 R02 本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。 本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。 batch-20261015-advance-20261117-r02-ratio-test-series-foundations · 2026-10-15 · 100 題 欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。 [欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json) ## 1. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":1,"ratio":[1,7],"shift":1,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(1, 1/7, 1, 否, a_n=1*(1/7)^n/(n+1), abs_ratio=1/7*(n+1)/(n+2), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(1, 1/7, 1, 否, a_n=1*(1/7)^n/(n+1), abs_ratio=1/7*(n+1)/(n+2), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0001/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 2. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":2,"ratio":[2,8],"shift":2,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(2, 1/4, 2, 是, a_n=2*(-1)^n*(1/4)^n/(n+2), abs_ratio=1/4*(n+2)/(n+3), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(2, 1/4, 2, 是, a_n=2*(-1)^n*(1/4)^n/(n+2), abs_ratio=1/4*(n+2)/(n+3), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0002/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 3. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":3,"ratio":[3,9],"shift":3,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(3, 1/3, 3, 否, a_n=3*(1/3)^n/(n+3), abs_ratio=1/3*(n+3)/(n+4), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(3, 1/3, 3, 否, a_n=3*(1/3)^n/(n+3), abs_ratio=1/3*(n+3)/(n+4), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0003/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 4. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":4,"ratio":[4,10],"shift":4,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(4, 2/5, 4, 是, a_n=4*(-1)^n*(2/5)^n/(n+4), abs_ratio=2/5*(n+4)/(n+5), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(4, 2/5, 4, 是, a_n=4*(-1)^n*(2/5)^n/(n+4), abs_ratio=2/5*(n+4)/(n+5), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0004/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 5. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":5,"ratio":[5,11],"shift":5,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(5, 5/11, 5, 否, a_n=5*(5/11)^n/(n+5), abs_ratio=5/11*(n+5)/(n+6), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(5, 5/11, 5, 否, a_n=5*(5/11)^n/(n+5), abs_ratio=5/11*(n+5)/(n+6), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0005/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 6. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":6,"ratio":[1,7],"shift":6,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(6, 1/7, 6, 是, a_n=6*(-1)^n*(1/7)^n/(n+6), abs_ratio=1/7*(n+6)/(n+7), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(6, 1/7, 6, 是, a_n=6*(-1)^n*(1/7)^n/(n+6), abs_ratio=1/7*(n+6)/(n+7), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0006/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 7. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":7,"ratio":[2,8],"shift":7,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(7, 1/4, 7, 否, a_n=7*(1/4)^n/(n+7), abs_ratio=1/4*(n+7)/(n+8), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(7, 1/4, 7, 否, a_n=7*(1/4)^n/(n+7), abs_ratio=1/4*(n+7)/(n+8), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0007/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 8. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":8,"ratio":[3,9],"shift":8,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(8, 1/3, 8, 是, a_n=8*(-1)^n*(1/3)^n/(n+8), abs_ratio=1/3*(n+8)/(n+9), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(8, 1/3, 8, 是, a_n=8*(-1)^n*(1/3)^n/(n+8), abs_ratio=1/3*(n+8)/(n+9), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0008/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 9. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":9,"ratio":[4,10],"shift":9,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(9, 2/5, 9, 否, a_n=9*(2/5)^n/(n+9), abs_ratio=2/5*(n+9)/(n+10), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(9, 2/5, 9, 否, a_n=9*(2/5)^n/(n+9), abs_ratio=2/5*(n+9)/(n+10), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0009/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 10. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":10,"ratio":[5,11],"shift":10,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(10, 5/11, 10, 是, a_n=10*(-1)^n*(5/11)^n/(n+10), abs_ratio=5/11*(n+10)/(n+11), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(10, 5/11, 10, 是, a_n=10*(-1)^n*(5/11)^n/(n+10), abs_ratio=5/11*(n+10)/(n+11), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0010/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 11. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":11,"ratio":[1,7],"shift":11,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(11, 1/7, 11, 否, a_n=11*(1/7)^n/(n+11), abs_ratio=1/7*(n+11)/(n+12), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(11, 1/7, 11, 否, a_n=11*(1/7)^n/(n+11), abs_ratio=1/7*(n+11)/(n+12), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0011/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 12. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":12,"ratio":[2,8],"shift":12,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(12, 1/4, 12, 是, a_n=12*(-1)^n*(1/4)^n/(n+12), abs_ratio=1/4*(n+12)/(n+13), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(12, 1/4, 12, 是, a_n=12*(-1)^n*(1/4)^n/(n+12), abs_ratio=1/4*(n+12)/(n+13), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0012/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 13. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":13,"ratio":[3,9],"shift":13,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(13, 1/3, 13, 否, a_n=13*(1/3)^n/(n+13), abs_ratio=1/3*(n+13)/(n+14), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(13, 1/3, 13, 否, a_n=13*(1/3)^n/(n+13), abs_ratio=1/3*(n+13)/(n+14), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0013/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 14. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":14,"ratio":[4,10],"shift":14,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(14, 2/5, 14, 是, a_n=14*(-1)^n*(2/5)^n/(n+14), abs_ratio=2/5*(n+14)/(n+15), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(14, 2/5, 14, 是, a_n=14*(-1)^n*(2/5)^n/(n+14), abs_ratio=2/5*(n+14)/(n+15), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0014/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 15. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":15,"ratio":[5,11],"shift":15,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(15, 5/11, 15, 否, a_n=15*(5/11)^n/(n+15), abs_ratio=5/11*(n+15)/(n+16), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(15, 5/11, 15, 否, a_n=15*(5/11)^n/(n+15), abs_ratio=5/11*(n+15)/(n+16), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0015/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 16. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":16,"ratio":[1,7],"shift":16,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(16, 1/7, 16, 是, a_n=16*(-1)^n*(1/7)^n/(n+16), abs_ratio=1/7*(n+16)/(n+17), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(16, 1/7, 16, 是, a_n=16*(-1)^n*(1/7)^n/(n+16), abs_ratio=1/7*(n+16)/(n+17), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0016/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 17. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":17,"ratio":[2,8],"shift":17,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(17, 1/4, 17, 否, a_n=17*(1/4)^n/(n+17), abs_ratio=1/4*(n+17)/(n+18), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(17, 1/4, 17, 否, a_n=17*(1/4)^n/(n+17), abs_ratio=1/4*(n+17)/(n+18), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0017/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 18. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":18,"ratio":[3,9],"shift":18,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(18, 1/3, 18, 是, a_n=18*(-1)^n*(1/3)^n/(n+18), abs_ratio=1/3*(n+18)/(n+19), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(18, 1/3, 18, 是, a_n=18*(-1)^n*(1/3)^n/(n+18), abs_ratio=1/3*(n+18)/(n+19), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0018/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 19. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":19,"ratio":[4,10],"shift":19,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(19, 2/5, 19, 否, a_n=19*(2/5)^n/(n+19), abs_ratio=2/5*(n+19)/(n+20), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(19, 2/5, 19, 否, a_n=19*(2/5)^n/(n+19), abs_ratio=2/5*(n+19)/(n+20), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0019/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 20. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":20,"ratio":[5,11],"shift":20,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(20, 5/11, 20, 是, a_n=20*(-1)^n*(5/11)^n/(n+20), abs_ratio=5/11*(n+20)/(n+21), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(20, 5/11, 20, 是, a_n=20*(-1)^n*(5/11)^n/(n+20), abs_ratio=5/11*(n+20)/(n+21), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0020/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 21. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":21,"ratio":[1,7],"shift":21,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(21, 1/7, 21, 否, a_n=21*(1/7)^n/(n+21), abs_ratio=1/7*(n+21)/(n+22), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(21, 1/7, 21, 否, a_n=21*(1/7)^n/(n+21), abs_ratio=1/7*(n+21)/(n+22), limit=1/7, 4/7, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0021/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 22. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":22,"ratio":[2,8],"shift":22,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(22, 1/4, 22, 是, a_n=22*(-1)^n*(1/4)^n/(n+22), abs_ratio=1/4*(n+22)/(n+23), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(22, 1/4, 22, 是, a_n=22*(-1)^n*(1/4)^n/(n+22), abs_ratio=1/4*(n+22)/(n+23), limit=1/4, 5/8, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0022/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 23. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":23,"ratio":[3,9],"shift":23,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(23, 1/3, 23, 否, a_n=23*(1/3)^n/(n+23), abs_ratio=1/3*(n+23)/(n+24), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(23, 1/3, 23, 否, a_n=23*(1/3)^n/(n+23), abs_ratio=1/3*(n+23)/(n+24), limit=1/3, 2/3, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0023/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 24. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":24,"ratio":[4,10],"shift":24,"alternating":true}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(24, 2/5, 24, 是, a_n=24*(-1)^n*(2/5)^n/(n+24), abs_ratio=2/5*(n+24)/(n+25), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(24, 2/5, 24, 是, a_n=24*(-1)^n*(2/5)^n/(n+24), abs_ratio=2/5*(n+24)/(n+25), limit=2/5, 7/10, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0024/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 25. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-nonconstant-convergent；資料={"scale":25,"ratio":[5,11],"shift":25,"alternating":false}。請依 [scale, ratio, shift, alternating, term_description, absolute_ratio_description, limit_description, dominating_ratio, ratio_bracket, domination_statement, conclusion] 欄序回答。 ratio 是完整有理數 r；現行追加式答案把 A*(r)^n/(n+shift) 的完整底數加上括號，不是 p/(q^n)。原先未加括號的顯示保留於歷史批次。 查看答案與解釋(25, 5/11, 25, 否, a_n=25*(5/11)^n/(n+25), abs_ratio=5/11*(n+25)/(n+26), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent) Wave6第十一層核對=(25, 5/11, 25, 否, a_n=25*(5/11)^n/(n+25), abs_ratio=5/11*(n+25)/(n+26), limit=5/11, 8/11, 是, eventual_geometric_domination, absolutely_convergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0025/) 級數比值判別法 · qf-mod48-ratio-nonconstant-convergent · candidate ## 26. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":1,"ratio":[6,3],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(1, 2, 否, 2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(1, 2, 否, 2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0026/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 27. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":2,"ratio":[7,4],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(2, 7/4, 是, 7/4, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(2, 7/4, 是, 7/4, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0027/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 28. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":3,"ratio":[8,5],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(3, 8/5, 否, 8/5, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(3, 8/5, 否, 8/5, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0028/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 29. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":4,"ratio":[9,6],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(4, 3/2, 是, 3/2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(4, 3/2, 是, 3/2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0029/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 30. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":5,"ratio":[10,7],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(5, 10/7, 否, 10/7, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(5, 10/7, 否, 10/7, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0030/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 31. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":6,"ratio":[6,3],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(6, 2, 是, 2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(6, 2, 是, 2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0031/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 32. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":7,"ratio":[7,4],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(7, 7/4, 否, 7/4, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(7, 7/4, 否, 7/4, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0032/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 33. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":8,"ratio":[8,5],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(8, 8/5, 是, 8/5, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(8, 8/5, 是, 8/5, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0033/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 34. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":9,"ratio":[9,6],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(9, 3/2, 否, 3/2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(9, 3/2, 否, 3/2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0034/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 35. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":10,"ratio":[10,7],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(10, 10/7, 是, 10/7, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(10, 10/7, 是, 10/7, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0035/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 36. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":11,"ratio":[6,3],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(11, 2, 否, 2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(11, 2, 否, 2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0036/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 37. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":12,"ratio":[7,4],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(12, 7/4, 是, 7/4, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(12, 7/4, 是, 7/4, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0037/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 38. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":13,"ratio":[8,5],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(13, 8/5, 否, 8/5, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(13, 8/5, 否, 8/5, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0038/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 39. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":14,"ratio":[9,6],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(14, 3/2, 是, 3/2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(14, 3/2, 是, 3/2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0039/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 40. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":15,"ratio":[10,7],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(15, 10/7, 否, 10/7, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(15, 10/7, 否, 10/7, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0040/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 41. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":16,"ratio":[6,3],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(16, 2, 是, 2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(16, 2, 是, 2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0041/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 42. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":17,"ratio":[7,4],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(17, 7/4, 否, 7/4, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(17, 7/4, 否, 7/4, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0042/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 43. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":18,"ratio":[8,5],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(18, 8/5, 是, 8/5, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(18, 8/5, 是, 8/5, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0043/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 44. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":19,"ratio":[9,6],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(19, 3/2, 否, 3/2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(19, 3/2, 否, 3/2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0044/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 45. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":20,"ratio":[10,7],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(20, 10/7, 是, 10/7, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(20, 10/7, 是, 10/7, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0045/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 46. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":21,"ratio":[6,3],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(21, 2, 否, 2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(21, 2, 否, 2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0046/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 47. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":22,"ratio":[7,4],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(22, 7/4, 是, 7/4, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(22, 7/4, 是, 7/4, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0047/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 48. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":23,"ratio":[8,5],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(23, 8/5, 否, 8/5, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(23, 8/5, 否, 8/5, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0048/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 49. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":24,"ratio":[9,6],"alternating":true}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(24, 3/2, 是, 3/2, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(24, 3/2, 是, 3/2, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0049/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 50. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-geometric-divergent；資料={"scale":25,"ratio":[10,7],"alternating":false}。請依 [scale, ratio, alternating, absolute_ratio, term_test_obstruction, conclusion] 欄序回答。 查看答案與解釋(25, 10/7, 否, 10/7, terms_do_not_tend_to_zero, divergent) Wave6第十一層核對=(25, 10/7, 否, 10/7, terms_do_not_tend_to_zero, divergent)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0050/) 級數比值判別法 · qf-mod48-ratio-geometric-divergent · candidate ## 51. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":1}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 1, abs_ratio=((n+1)/(n+2))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 1, abs_ratio=((n+1)/(n+2))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0051/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 52. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":2}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 2, abs_ratio=((n+2)/(n+3))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 2, abs_ratio=((n+2)/(n+3))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0052/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 53. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":3}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 3, abs_ratio=((n+3)/(n+4))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 3, abs_ratio=((n+3)/(n+4))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0053/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 54. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":4}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 4, abs_ratio=((n+4)/(n+5))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 4, abs_ratio=((n+4)/(n+5))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0054/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 55. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":5}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 5, abs_ratio=((n+5)/(n+6))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 5, abs_ratio=((n+5)/(n+6))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0055/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 56. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":6}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 6, abs_ratio=((n+6)/(n+7))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 6, abs_ratio=((n+6)/(n+7))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0056/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 57. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":7}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 7, abs_ratio=((n+7)/(n+8))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 7, abs_ratio=((n+7)/(n+8))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0057/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 58. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":8}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 8, abs_ratio=((n+8)/(n+9))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 8, abs_ratio=((n+8)/(n+9))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0058/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 59. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":9}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 9, abs_ratio=((n+9)/(n+10))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 9, abs_ratio=((n+9)/(n+10))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0059/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 60. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":10}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 10, abs_ratio=((n+10)/(n+11))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 10, abs_ratio=((n+10)/(n+11))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0060/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 61. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":11}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 11, abs_ratio=((n+11)/(n+12))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 11, abs_ratio=((n+11)/(n+12))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0061/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 62. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":12}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 12, abs_ratio=((n+12)/(n+13))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 12, abs_ratio=((n+12)/(n+13))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0062/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 63. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":13}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 13, abs_ratio=((n+13)/(n+14))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 13, abs_ratio=((n+13)/(n+14))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0063/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 64. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":14}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 14, abs_ratio=((n+14)/(n+15))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 14, abs_ratio=((n+14)/(n+15))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0064/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 65. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":15}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 15, abs_ratio=((n+15)/(n+16))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 15, abs_ratio=((n+15)/(n+16))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0065/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 66. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":16}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 16, abs_ratio=((n+16)/(n+17))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 16, abs_ratio=((n+16)/(n+17))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0066/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 67. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":17}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 17, abs_ratio=((n+17)/(n+18))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 17, abs_ratio=((n+17)/(n+18))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0067/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 68. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":18}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 18, abs_ratio=((n+18)/(n+19))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 18, abs_ratio=((n+18)/(n+19))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0068/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 69. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":19}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 19, abs_ratio=((n+19)/(n+20))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 19, abs_ratio=((n+19)/(n+20))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0069/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 70. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":20}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 20, abs_ratio=((n+20)/(n+21))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 20, abs_ratio=((n+20)/(n+21))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0070/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 71. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":21}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 21, abs_ratio=((n+21)/(n+22))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 21, abs_ratio=((n+21)/(n+22))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0071/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 72. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":22}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 22, abs_ratio=((n+22)/(n+23))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 22, abs_ratio=((n+22)/(n+23))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0072/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 73. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":23}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 23, abs_ratio=((n+23)/(n+24))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 23, abs_ratio=((n+23)/(n+24))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0073/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 74. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":2,"shift":24}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(2, 24, abs_ratio=((n+24)/(n+25))^2, 1, convergent_p_series, L_equal_one_is_inconclusive) Wave6第十一層核對=(2, 24, abs_ratio=((n+24)/(n+25))^2, 1, convergent_p_series, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0074/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 75. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-limit-one-inconclusive；資料={"exponent":1,"shift":25}。請依 [exponent, shift, absolute_ratio_description, ratio_limit, actual_series_status, ratio_test_scope] 欄序回答。 查看答案與解釋(1, 25, abs_ratio=((n+25)/(n+26))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive) Wave6第十一層核對=(1, 25, abs_ratio=((n+25)/(n+26))^1, 1, divergent_harmonic_tail, L_equal_one_is_inconclusive)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0075/) 級數比值判別法 · qf-mod48-ratio-limit-one-inconclusive · candidate ## 76. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":2}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 2, a_n=2_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 2, a_n=2_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0076/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 77. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"finite_samples","scale":3}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(finite_samples, 3, ratios_equal_1/2_for_n<=3_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否) Wave6第十一層核對=(finite_samples, 3, ratios_equal_1/2_for_n<=3_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0077/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 78. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"missing_limit","scale":4}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(missing_limit, 4, absolute_ratios_alternate_between_1/5_and_5, limit_does_not_exist, 否) Wave6第十一層核對=(missing_limit, 4, absolute_ratios_alternate_between_1/5_and_5, limit_does_not_exist, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0078/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 79. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"signed_without_absolute","scale":5}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(signed_without_absolute, 5, a_n=(-1)^n*5*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否) Wave6第十一層核對=(signed_without_absolute, 5, a_n=(-1)^n*5*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0079/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 80. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":6}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 6, a_n=6_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 6, a_n=6_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0080/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 81. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"finite_samples","scale":7}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(finite_samples, 7, ratios_equal_1/2_for_n<=7_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否) Wave6第十一層核對=(finite_samples, 7, ratios_equal_1/2_for_n<=7_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0081/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 82. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"missing_limit","scale":8}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(missing_limit, 8, absolute_ratios_alternate_between_1/9_and_9, limit_does_not_exist, 否) Wave6第十一層核對=(missing_limit, 8, absolute_ratios_alternate_between_1/9_and_9, limit_does_not_exist, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0082/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 83. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"signed_without_absolute","scale":9}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(signed_without_absolute, 9, a_n=(-1)^n*9*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否) Wave6第十一層核對=(signed_without_absolute, 9, a_n=(-1)^n*9*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0083/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 84. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":10}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 10, a_n=10_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 10, a_n=10_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0084/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 85. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"finite_samples","scale":11}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(finite_samples, 11, ratios_equal_1/2_for_n<=11_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否) Wave6第十一層核對=(finite_samples, 11, ratios_equal_1/2_for_n<=11_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0085/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 86. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"missing_limit","scale":12}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(missing_limit, 12, absolute_ratios_alternate_between_1/13_and_13, limit_does_not_exist, 否) Wave6第十一層核對=(missing_limit, 12, absolute_ratios_alternate_between_1/13_and_13, limit_does_not_exist, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0086/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 87. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"signed_without_absolute","scale":13}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(signed_without_absolute, 13, a_n=(-1)^n*13*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否) Wave6第十一層核對=(signed_without_absolute, 13, a_n=(-1)^n*13*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0087/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 88. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":14}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 14, a_n=14_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 14, a_n=14_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0088/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 89. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"finite_samples","scale":15}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(finite_samples, 15, ratios_equal_1/2_for_n<=15_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否) Wave6第十一層核對=(finite_samples, 15, ratios_equal_1/2_for_n<=15_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0089/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 90. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"missing_limit","scale":16}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(missing_limit, 16, absolute_ratios_alternate_between_1/17_and_17, limit_does_not_exist, 否) Wave6第十一層核對=(missing_limit, 16, absolute_ratios_alternate_between_1/17_and_17, limit_does_not_exist, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0090/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 91. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"signed_without_absolute","scale":17}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(signed_without_absolute, 17, a_n=(-1)^n*17*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否) Wave6第十一層核對=(signed_without_absolute, 17, a_n=(-1)^n*17*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0091/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 92. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":18}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 18, a_n=18_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 18, a_n=18_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0092/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 93. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"finite_samples","scale":19}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(finite_samples, 19, ratios_equal_1/2_for_n<=19_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否) Wave6第十一層核對=(finite_samples, 19, ratios_equal_1/2_for_n<=19_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0093/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 94. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"missing_limit","scale":20}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(missing_limit, 20, absolute_ratios_alternate_between_1/21_and_21, limit_does_not_exist, 否) Wave6第十一層核對=(missing_limit, 20, absolute_ratios_alternate_between_1/21_and_21, limit_does_not_exist, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0094/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 95. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"signed_without_absolute","scale":21}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(signed_without_absolute, 21, a_n=(-1)^n*21*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否) Wave6第十一層核對=(signed_without_absolute, 21, a_n=(-1)^n*21*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0095/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 96. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":22}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 22, a_n=22_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 22, a_n=22_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0096/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 97. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"finite_samples","scale":23}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(finite_samples, 23, ratios_equal_1/2_for_n<=23_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否) Wave6第十一層核對=(finite_samples, 23, ratios_equal_1/2_for_n<=23_then_equal_2, finite_prefix_cannot_determine_ratio_limit, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0097/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 98. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"missing_limit","scale":24}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(missing_limit, 24, absolute_ratios_alternate_between_1/25_and_25, limit_does_not_exist, 否) Wave6第十一層核對=(missing_limit, 24, absolute_ratios_alternate_between_1/25_and_25, limit_does_not_exist, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0098/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 99. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"signed_without_absolute","scale":25}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(signed_without_absolute, 25, a_n=(-1)^n*25*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否) Wave6第十一層核對=(signed_without_absolute, 25, a_n=(-1)^n*25*(1/2)^n, raw_ratio=-1/2_but_absolute_ratio=1/2, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0099/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate ## 100. 級數比值判別法 eventually nonzero、absolute successive ratio與limit存在責任。操作=ratio-test-boundary；資料={"mode":"zero_terms","scale":26}。請依 [mode, scale, series_or_ratio_description, scope_or_obstruction, method_application_claim] 欄序回答。 最後否保留ratio-test方法前提，不是在signed_without_absolute例宣稱原級數發散；該例的absolute ratio仍小於1。 查看答案與解釋(zero_terms, 26, a_n=26_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否) Wave6第十一層核對=(zero_terms, 26, a_n=26_for_odd_n_and_0_for_even_n, successive_ratio_has_zero_denominators_infinitely_often, 否)；依完整明示資料獨立重算；L<1絕對收斂、L>1發散、L=1無結論；有限ratio samples、zero denominators或raw signed ratio不能冒充假設。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261015-advance-20261117-r02-ratio-test-series-0100/) 級數比值判別法 · qf-mod48-ratio-test-boundary · candidate [原始題目 JSONL](https://ocme.evemisslab.com/data/questions/revisions/ocme-notation-guard-revision-20261018/batch-20261015-advance-20261117-r02-ratio-test-series-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/revisions/ocme-notation-guard-revision-20261018/batch-20261015-advance-20261117-r02-ratio-test-series-foundations/validation-report.json)