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