Question

Is there a general method to find $\sum_{i = 1}^{N} i^a$

score 119 · Answer 1 · 2019-10-07 09:19:19

119

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Is there a general method to find $\sum_{i = 1}^{N} i^a$

Published on 07 Oct 2019 - 9:19

score 126 · Answer 2 · 2019-10-17 11:32:42

126

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On $\int_0^1\frac{\ln(1-e^{\pi i/3}x)}{e^{-\pi i/3}-x}\ln^3xdx$ and its generalization

Published on 17 Oct 2019 - 11:32

score 505 · Answer 3 · 2019-11-26 03:51:14

505

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An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$

Published on 26 Nov 2019 - 3:51

score 567 · Answer 4 · 2020-01-26 07:58:32

567

Views

Closed form for the skew-harmonic sum $\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}$

Published on 26 Jan 2020 - 7:58

score 436 · Answer 5 · 2020-01-30 22:15:26

436

Views

Evaluating $\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx$

Published on 30 Jan 2020 - 22:15

score 850 · Answer 6 · 2020-02-19 05:47:28

850

Views

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

Published on 19 Feb 2020 - 5:47

score 280 · Answer 7 · 2020-05-07 12:59:55

280

Views

Binomial Euler sums: Evaluate $\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$

Published on 07 May 2020 - 12:59

score 97 · Answer 8 · 2020-05-19 17:44:46

97

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Formula for the general case of $\sum_{n=1}^\infty \frac{H_n}{n^2} = 2\zeta(3)$

Published on 19 May 2020 - 17:44

score 72 · Answer 9 · 2020-05-28 16:55:39

72

Views

Nonlinear alternating Euler sums

Published on 28 May 2020 - 16:55

score 166 · Answer 10 · 2020-05-30 10:29:23

166

Views

Quadratic Euler Sum $\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}\left(\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}\right)^2$

Published on 30 May 2020 - 10:29

score 274 · Answer 11 · 2020-06-11 09:14:00

274

Views

Quadratic Euler sums $\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \widetilde H(n)^3}{2 n+1}$

Published on 11 Jun 2020 - 9:14

score 320 · Answer 12 · 2017-01-24 21:40:37

320

Views

A cubic nonlinear Euler sum

Published on 24 Jan 2017 - 21:40

score 8.6k · Answer 13 · 2017-03-03 01:04:13

8.6k

Views

Calculating alternating Euler sums of odd powers

Published on 03 Mar 2017 - 1:04

score 70 · Answer 14 · 2023-02-23 19:48:42

70

Views

Can the generating function $\sum_{n = 1}^{\infty} H_n^k z^n$ for $k \in \mathbb{N}$ generally be determined by special functions?

Published on 23 Feb 2023 - 19:48

score 97 · Answer 15 · 2023-03-23 17:37:05

97

Views

Closed form of series involving harmonic numbers

Published on 23 Mar 2023 - 17:37

List Question

Is there a general method to find $\sum_{i = 1}^{N} i^a$

On $\int_0^1\frac{\ln(1-e^{\pi i/3}x)}{e^{-\pi i/3}-x}\ln^3xdx$ and its generalization

An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$

Closed form for the skew-harmonic sum $\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}$

Evaluating $\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx$

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

Binomial Euler sums: Evaluate $\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$

Formula for the general case of $\sum_{n=1}^\infty \frac{H_n}{n^2} = 2\zeta(3)$

Nonlinear alternating Euler sums

Quadratic Euler Sum $\sum_{n=1}^\infty \frac{(\pm1)^{n-1}}{n^2}\left(\sum_{m=1}^n \frac{(-1)^{m-1}}{2m-1}\right)^2$

Quadratic Euler sums $\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \widetilde H(n)^3}{2 n+1}$

A cubic nonlinear Euler sum

Calculating alternating Euler sums of odd powers

Can the generating function $\sum_{n = 1}^{\infty} H_n^k z^n$ for $k \in \mathbb{N}$ generally be determined by special functions?

Closed form of series involving harmonic numbers

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