Question

Sum $\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$

score 815 · Answer 1 · 2026-03-26 01:07:57

815

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Sum $\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$

Published on 26 Mar 2026 - 1:07

score 480 · Answer 2 · 2014-09-04 17:52:35

480

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Another way of expressing $\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$

Published on 04 Sep 2014 - 17:52

score 107 · Answer 3 · 2026-03-25 07:41:19

107

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Function related to Harmonic numbers, the Pascal triangle, Logarithmic integral and the Polylogarithm.

Published on 25 Mar 2026 - 7:41

score 4.1k · Answer 4 · 2014-09-16 13:44:14

4.1k

Views

Sum of Squares of Harmonic Numbers

Published on 16 Sep 2014 - 13:44

score 940 · Answer 5 · 2014-09-16 18:04:18

940

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Integral $\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$

Published on 16 Sep 2014 - 18:04

score 1.6k · Answer 6 · 2014-09-17 12:43:50

1.6k

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Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$

Published on 17 Sep 2014 - 12:43

score 182 · Answer 7 · 2014-09-22 18:53:41

182

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Can $\sum_1^n 1/k$ be arranged so that it is an integer for infinitely many $n?$

Published on 22 Sep 2014 - 18:53

score 1.8k · Answer 8 · 2014-09-24 07:27:30

1.8k

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Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

Published on 24 Sep 2014 - 7:27

score 219 · Answer 9 · 2014-09-30 13:33:25

219

Views

A cute limit $\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$

Published on 30 Sep 2014 - 13:33

score 484 · Answer 10 · 2026-03-25 11:16:31

484

Views

Possible new definition of Gamma (Euler-Mascheroni Constant): $\lim_{x \to 0} (-\ln ( \sqrt[x]{x!} )) = \gamma$

Published on 25 Mar 2026 - 11:16

score 661 · Answer 11 · 2014-10-02 20:17:17

661

Views

Infinite Sum of the Reciprocal of the Inverse Harmonic Number Function

Published on 02 Oct 2014 - 20:17

score 622 · Answer 12 · 2014-10-04 14:09:41

622

Views

Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$

Published on 04 Oct 2014 - 14:09

score 231 · Answer 13 · 2026-03-25 07:43:50

231

Views

Double series of Harmonic Numbers

Published on 25 Mar 2026 - 7:43

score 139 · Answer 14 · 2014-10-07 14:56:19

139

Views

Is there any way to evaluate $e^{H_n} = … + O(\frac{1}{n})$, where $H_n$ is $n$-th harmonic number?

Published on 07 Oct 2014 - 14:56

score 1.1k · Answer 15 · 2026-03-26 02:47:40

1.1k

Views

Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$

Published on 26 Mar 2026 - 2:47

List Question

Sum $\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$

Another way of expressing $\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$

Function related to Harmonic numbers, the Pascal triangle, Logarithmic integral and the Polylogarithm.

Sum of Squares of Harmonic Numbers

Integral $\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$

Closed-form of $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\Psi_3(n+1)=-\int_0^1\frac{\ln(1+x)\ln^3 x}{1-x}\,dx$

Can $\sum_1^n 1/k$ be arranged so that it is an integer for infinitely many $n?$

Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

A cute limit $\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$

Possible new definition of Gamma (Euler-Mascheroni Constant): $\lim_{x \to 0} (-\ln ( \sqrt[x]{x!} )) = \gamma$

Infinite Sum of the Reciprocal of the Inverse Harmonic Number Function

Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$

Double series of Harmonic Numbers

Is there any way to evaluate $e^{H_n} = … + O(\frac{1}{n})$, where $H_n$ is $n$-th harmonic number?

Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$

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