Question

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

score 173 · Answer 1 · 2026-05-10 15:02:31

173

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

Published on 10 May 2026 - 15:02

score 78 · Answer 2 · 2026-05-10 14:49:01

78

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How to prove the following equality involving sums

Published on 10 May 2026 - 14:49

score 112 · Answer 3 · 2026-05-10 15:02:05

112

Views

Evaluation of a series involving the harmonic number

Published on 10 May 2026 - 15:02

score 154 · Answer 4 · 2026-05-10 13:49:07

154

Views

Alternating series combining harmonic number and zeta values

Published on 10 May 2026 - 13:49

score 414 · Answer 5 · 2026-05-10 13:51:03

414

Views

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 10 May 2026 - 13:51

score 87 · Answer 6 · 2026-05-10 13:51:20

87

Views

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 10 May 2026 - 13:51

score 836 · Answer 7 · 2026-05-10 15:02:29

836

Views

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

Published on 10 May 2026 - 15:02

score 295 · Answer 8 · 2026-05-10 13:49:18

295

Views

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

Published on 10 May 2026 - 13:49

score 484 · Answer 9 · 2026-05-10 14:45:25

484

Views

A peculiar Euler sum

Published on 10 May 2026 - 14:45

score 682 · Answer 10 · 2026-05-07 10:14:09

682

Views

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

Published on 07 May 2026 - 10:14

score 128 · Answer 11 · 2026-05-10 13:45:10

128

Views

famous Euler sum

Published on 10 May 2026 - 13:45

score 255 · Answer 12 · 2026-05-10 13:50:44

255

Views

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

Published on 10 May 2026 - 13:50

score 120 · Answer 13 · 2026-05-10 13:51:33

120

Views

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

Published on 10 May 2026 - 13:51

score 466 · Answer 14 · 2026-05-10 13:47:53

466

Views

A difficult logarithmic integral and its relation to alternating Euler Sums

Published on 10 May 2026 - 13:47

score 153 · Answer 15 · 2026-05-10 13:45:06

153

Views

Two alternating Euler sums

Published on 10 May 2026 - 13:45

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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