# Exact Partial Sums, Geometric and Telescoping Tails, Harmonic Boundaries

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261014-advance-20261103-r03-infinite-series-foundations · 2026-10-14 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/2, -1/3], 1/6, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0001/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 2. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/3, -1/4], 1/12, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0002/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 3. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/4, -1/5], 1/20, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0003/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 4. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/5, -1/6], 1/30, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0004/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 5. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/6, -1/7], 1/42, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0005/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 6. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/7, -1/8], 1/56, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0006/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 7. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/8, -1/9], 1/72, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0007/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 8. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/9, -1/10], 1/90, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0008/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 9. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/10, -1/11], 1/110, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0009/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 10. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/11, -1/12], 1/132, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0010/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 11. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/12, -1/13], 1/156, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0011/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 12. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/13, -1/14], 1/182, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0012/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 13. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/14, -1/15], 1/210, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0013/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 14. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/15, -1/16], 1/240, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0014/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 15. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/16, -1/17], 1/272, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0015/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 16. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/17, -1/18], 1/306, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0016/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 17. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/18, -1/19], 1/342, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0017/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 18. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/19, -1/20], 1/380, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0018/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 19. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/20, -1/21], 1/420, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0019/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 20. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/21, -1/22], 1/462, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0020/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 21. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/22, -1/23], 1/506, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0021/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 22. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/23, -1/24], 1/552, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0022/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 23. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/24, -1/25], 1/600, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0023/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 24. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/25, -1/26], 1/650, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0024/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 25. A Finite Partial Sum Is Not an Infinite-Series Limit

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.

View answer and explanation([1/26, -1/27], 1/702, 2, finite_partial_sum_only)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0025/)

Infinite series · qf-mod34-series-partial-sum · candidate





## 26. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(1/2, 1/2, 1, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0026/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 27. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(2/3, 1/2, 4/3, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0027/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 28. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(3/4, 1/2, 3/2, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0028/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 29. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(4/5, 1/2, 8/5, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0029/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 30. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(5/6, 1/2, 5/3, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0030/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 31. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(6/7, 1/2, 12/7, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0031/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 32. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(7/8, 1/2, 7/4, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0032/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 33. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(8/9, 1/2, 16/9, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0033/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 34. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(9/10, 1/2, 9/5, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0034/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 35. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(10/11, 1/2, 20/11, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0035/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 36. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(11/12, 1/2, 11/6, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0036/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 37. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(12/13, 1/2, 24/13, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0037/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 38. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(13/14, 1/2, 13/7, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0038/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 39. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(14/15, 1/2, 28/15, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0039/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 40. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(15/16, 1/2, 15/8, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0040/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 41. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(16/17, 1/2, 32/17, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0041/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 42. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(17/18, 1/2, 17/9, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0042/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 43. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(18/19, 1/2, 36/19, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0043/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 44. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(19/20, 1/2, 19/10, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0044/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 45. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(20/21, 1/2, 40/21, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0045/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 46. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(21/22, 1/2, 21/11, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0046/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 47. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(22/23, 1/2, 44/23, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0047/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 48. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(23/24, 1/2, 23/12, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0048/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 49. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(24/25, 1/2, 48/25, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0049/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 50. The Geometric-Series Formula Retains Its Absolute-Ratio Premise

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).

View answer and explanation(25/26, 1/2, 25/13, absolute_ratio_below_one_symbolic_tail)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0050/)

Infinite series · qf-mod34-series-geometric-sum · candidate





## 51. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(1, 2, 1, 1/6, 1/3, 1/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0051/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 52. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(2, 2, 2, 1/2, 2/4, 2/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0052/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 53. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(3, 2, 3, 9/10, 3/5, 3/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0053/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 54. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(4, 2, 4, 4/3, 4/6, 4/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0054/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 55. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(5, 2, 5, 25/14, 5/7, 5/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0055/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 56. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(6, 2, 6, 9/4, 6/8, 6/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0056/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 57. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(7, 2, 7, 49/18, 7/9, 7/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0057/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 58. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(8, 2, 8, 16/5, 8/10, 8/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0058/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 59. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(9, 2, 9, 81/22, 9/11, 9/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0059/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 60. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(10, 2, 10, 25/6, 10/12, 10/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0060/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 61. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(11, 2, 11, 121/26, 11/13, 11/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0061/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 62. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(12, 2, 12, 36/7, 12/14, 12/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0062/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 63. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(13, 2, 13, 169/30, 13/15, 13/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0063/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 64. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(14, 2, 14, 49/8, 14/16, 14/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0064/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 65. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(15, 2, 15, 225/34, 15/17, 15/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0065/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 66. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(16, 2, 16, 64/9, 16/18, 16/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0066/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 67. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(17, 2, 17, 289/38, 17/19, 17/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0067/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 68. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0068/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 69. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(19, 2, 19, 361/42, 19/21, 19/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0069/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 70. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(20, 2, 20, 100/11, 20/22, 20/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0070/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 71. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(21, 2, 21, 441/46, 21/23, 21/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0071/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 72. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(22, 2, 22, 121/12, 22/24, 22/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0072/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 73. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(23, 2, 23, 529/50, 23/25, 23/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0073/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 74. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(24, 2, 24, 144/13, 24/26, 24/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0074/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 75. Retain Both the Telescoping Remainder and the Source Ratio Text

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.

View answer and explanation(25, 2, 25, 625/54, 25/27, 25/2, exact_telescoping_remainder)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0075/)

Infinite series · qf-mod34-series-telescoping-sum · candidate





## 76. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(1, 5, [1, 1/2, 1/3, 1/4, 1/5], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0076/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 77. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(2, 5, [1/2, 1/3, 1/4, 1/5, 1/6], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0077/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 78. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(3, 5, [1/3, 1/4, 1/5, 1/6, 1/7], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0078/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 79. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(4, 5, [1/4, 1/5, 1/6, 1/7, 1/8], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0079/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 80. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(5, 5, [1/5, 1/6, 1/7, 1/8, 1/9], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0080/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 81. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(6, 5, [1/6, 1/7, 1/8, 1/9, 1/10], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0081/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 82. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(7, 5, [1/7, 1/8, 1/9, 1/10, 1/11], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0082/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 83. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(8, 5, [1/8, 1/9, 1/10, 1/11, 1/12], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0083/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 84. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(9, 5, [1/9, 1/10, 1/11, 1/12, 1/13], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0084/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 85. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(10, 5, [1/10, 1/11, 1/12, 1/13, 1/14], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0085/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 86. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(11, 5, [1/11, 1/12, 1/13, 1/14, 1/15], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0086/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 87. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(12, 5, [1/12, 1/13, 1/14, 1/15, 1/16], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0087/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 88. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(13, 5, [1/13, 1/14, 1/15, 1/16, 1/17], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0088/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 89. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(14, 5, [1/14, 1/15, 1/16, 1/17, 1/18], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0089/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 90. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(15, 5, [1/15, 1/16, 1/17, 1/18, 1/19], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0090/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 91. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(16, 5, [1/16, 1/17, 1/18, 1/19, 1/20], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0091/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 92. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(17, 5, [1/17, 1/18, 1/19, 1/20, 1/21], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0092/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 93. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(18, 5, [1/18, 1/19, 1/20, 1/21, 1/22], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0093/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 94. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(19, 5, [1/19, 1/20, 1/21, 1/22, 1/23], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0094/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 95. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(20, 5, [1/20, 1/21, 1/22, 1/23, 1/24], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0095/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 96. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(21, 5, [1/21, 1/22, 1/23, 1/24, 1/25], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0096/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 97. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(22, 5, [1/22, 1/23, 1/24, 1/25, 1/26], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0097/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 98. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(23, 5, [1/23, 1/24, 1/25, 1/26, 1/27], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0098/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 99. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(24, 5, [1/24, 1/25, 1/26, 1/27, 1/28], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0099/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate





## 100. Terms Tending to Zero Do Not Make a Harmonic Tail Summable

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).

View answer and explanation(25, 5, [1/25, 1/26, 1/27, 1/28, 1/29], Yes, No, terms_tend_to_zero_but_shifted_harmonic_series_diverges)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261103-r03-infinite-series-0100/)

Infinite series · qf-mod34-series-term-zero-boundary · candidate



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