Question

Prove $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$

score 341 · Answer 1 · 2019-07-18 22:40:02

341

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Prove $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$

Published on 18 Jul 2019 - 22:40

score 378 · Answer 2 · 2019-07-20 08:09:35

378

Views

Closed form for $\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) x^k$

Published on 20 Jul 2019 - 8:09

score 649 · Answer 3 · 2019-07-21 01:08:38

649

Views

Resistant integral $\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$

Published on 21 Jul 2019 - 1:08

score 291 · Answer 4 · 2019-07-24 10:31:50

291

Views

The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$

Published on 24 Jul 2019 - 10:31

score 1.5k · Answer 5 · 2026-03-25 14:25:21

1.5k

Views

Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$

Published on 25 Mar 2026 - 14:25

score 411 · Answer 6 · 2019-07-28 18:37:51

411

Views

Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$

Published on 28 Jul 2019 - 18:37

score 255 · Answer 7 · 2019-07-29 02:02:54

255

Views

Compute $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^42^k{2k \choose k}}$

Published on 29 Jul 2019 - 2:02

score 382 · Answer 8 · 2026-03-25 16:02:51

382

Views

Challenging sum: Calculate $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^52^k{2k \choose k}}$

Published on 25 Mar 2026 - 16:02

score 140 · Answer 9 · 2019-07-31 19:32:33

140

Views

Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$

Published on 31 Jul 2019 - 19:32

score 421 · Answer 10 · 2019-08-01 20:01:01

421

Views

Compute $\sum_{n=1}^\infty\frac{H_n^4}{n^2}$ and $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}$

Published on 01 Aug 2019 - 20:01

score 500 · Answer 11 · 2019-08-02 06:01:38

500

Views

Compute $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}$ and $\sum_{n=1}^\infty\frac{H_n^2}{n^7}$

Published on 02 Aug 2019 - 6:01

score 298 · Answer 12 · 2019-08-02 21:22:47

298

Views

Is the result for $3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}$ known in the literature?

Published on 02 Aug 2019 - 21:22

score 223 · Answer 13 · 2019-08-03 21:42:01

223

Views

The limit $\lim_{r\to0}\frac1r\left(1-\binom{n}{r}^{-1}\right)$

Published on 03 Aug 2019 - 21:42

score 224 · Answer 14 · 2019-08-05 15:54:27

224

Views

Existence of Inverse Fourier Transform for a function

Published on 05 Aug 2019 - 15:54

score 1.1k · Answer 15 · 2019-08-05 21:55:33

1.1k

Views

Compute $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^3}$

Published on 05 Aug 2019 - 21:55

List Question

Prove $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)$

Closed form for $\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) x^k$

Resistant integral $\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$

The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$

Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$

Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$

Compute $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^42^k{2k \choose k}}$

Challenging sum: Calculate $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^52^k{2k \choose k}}$

Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$

Compute $\sum_{n=1}^\infty\frac{H_n^4}{n^2}$ and $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}$

Compute $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}$ and $\sum_{n=1}^\infty\frac{H_n^2}{n^7}$

Is the result for $3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}$ known in the literature?

The limit $\lim_{r\to0}\frac1r\left(1-\binom{n}{r}^{-1}\right)$

Existence of Inverse Fourier Transform for a function

Compute $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^3}$

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