Question

Sum identity over the reciprocal of central binomial coefficients using their partial fraction decompositions.

score 166 · Answer 1 · 2026-03-25 01:17:20

166

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Sum identity over the reciprocal of central binomial coefficients using their partial fraction decompositions.

Published on 25 Mar 2026 - 1:17

score 72 · Answer 2 · 2026-03-19 14:43:16

72

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How to prove the following equality involving sums

Published on 19 Mar 2026 - 14:43

score 107 · Answer 3 · 2026-03-24 23:46:11

107

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Evaluation of a series involving the harmonic number

Published on 24 Mar 2026 - 23:46

score 148 · Answer 4 · 2026-03-24 23:44:13

148

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Alternating series combining harmonic number and zeta values

Published on 24 Mar 2026 - 23:44

score 409 · Answer 5 · 2026-03-25 03:04:19

409

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Can $\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}$, with $H_n$ the $n$-th harmonic number, be written in terms of $\zeta$ values?

Published on 25 Mar 2026 - 3:04

score 82 · Answer 6 · 2026-03-25 01:18:39

82

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Closed form for $\sum\limits_{n=1}^{\infty}\frac{O_{n}^{(p)}}{(2n-1)^{q}}$, with $O_{n}^{(s)}=\sum\limits_{k=1}^n\frac1{(2k-1)^{s}}$

Published on 25 Mar 2026 - 1:18

score 831 · Answer 7 · 2026-03-25 01:17:20

831

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Evaluate $\int_0^1 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx$

Published on 25 Mar 2026 - 1:17

score 290 · Answer 8 · 2026-03-25 01:24:02

290

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Two Euler sums each containing the reciprocal of the central binomial coefficient

Published on 25 Mar 2026 - 1:24

score 463 · Answer 9 · 2026-03-25 01:13:11

463

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A peculiar Euler sum

Published on 25 Mar 2026 - 1:13

score 678 · Answer 10 · 2026-03-25 01:20:40

678

Views

How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?

Published on 25 Mar 2026 - 1:20

score 106 · Answer 11 · 2026-03-25 01:13:20

106

Views

famous Euler sum

Published on 25 Mar 2026 - 1:13

score 246 · Answer 12 · 2026-03-25 03:03:02

246

Views

On the alternating Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1}$

Published on 25 Mar 2026 - 3:03

score 111 · Answer 13 · 2026-03-19 20:48:05

111

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How to prove/disprove that $\sum _{k=1}^{\infty } \frac{(-1)^k \left(H_{k-1}+\log (k)+\gamma \right)}{k}= 0$

Published on 19 Mar 2026 - 20:48

score 456 · Answer 14 · 2026-03-25 02:59:23

456

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A difficult logarithmic integral and its relation to alternating Euler Sums

Published on 25 Mar 2026 - 2:59

score 141 · Answer 15 · 2026-03-25 02:56:41

141

Views

Two alternating Euler sums

Published on 25 Mar 2026 - 2:56

List Question

Sum identity over the reciprocal of central binomial coefficients using their partial fraction decompositions.

How to prove the following equality involving sums

Evaluation of a series involving the harmonic number

Alternating series combining harmonic number and zeta values

Can $\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}$, with $H_n$ the $n$-th harmonic number, be written in terms of $\zeta$ values?

Closed form for $\sum\limits_{n=1}^{\infty}\frac{O_{n}^{(p)}}{(2n-1)^{q}}$, with $O_{n}^{(s)}=\sum\limits_{k=1}^n\frac1{(2k-1)^{s}}$

Evaluate $\int_0^1 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx$

Two Euler sums each containing the reciprocal of the central binomial coefficient

A peculiar Euler sum

How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?

famous Euler sum

On the alternating Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1}$

How to prove/disprove that $\sum _{k=1}^{\infty } \frac{(-1)^k \left(H_{k-1}+\log (k)+\gamma \right)}{k}= 0$

A difficult logarithmic integral and its relation to alternating Euler Sums

Two alternating Euler sums

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