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$$\sum_{k=1}^{\infty} \frac{H_{k}^{(2)}}{2^kk} = \frac{5\zeta(3)}{8}$$

Can someone show this without making a generating function for

$$\sum_{k=1}^{\infty} \frac{x^kH_{k}^{(2)}}{k}$$

Since that is the route I took and successfully got the answer. I totally feel as if there’s a nicer way to go about it, as the answer is so simple in terms of the zeta function.

To be clear: $$H_{n}^{(m)} = \sum_{k=1}^{n} \frac{1}{k^m}$$

MichaelCatliMath
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  • No you obviously didn’t. It’s not really that hard to deduce from what I wrote that I don’t want any type of solution that starts with the step of considering that generating function. – MichaelCatliMath May 23 '20 at 20:13
  • It's not really hard to deduce that both sentences are semantically equivalent. Good luck. –  May 23 '20 at 20:48
  • All I was asking for was someone to approach the problem without the generating function - just out of curiosity. No need to be passive aggressive here, if you’re confused just tell me without being so rude. – MichaelCatliMath May 23 '20 at 20:52
  • Don't be, then. This is not a "do my work" site. Show your own effort, and with some luck, you may find support. –  May 23 '20 at 20:58
  • It’s not like my homework or anything, it’s just a problem I thought of and am interested in other people’s input. – MichaelCatliMath May 23 '20 at 21:58
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    You can derive identities like $$\sum _{n=1}^{\infty } \frac{1}{2^n n}\sum _{k=1}^n \frac{1}{k^2}=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty } \frac{1}{2^n n} \left(\frac{1}{k^2}-\frac{1}{(k+n)^2}\right)=\frac{\pi ^2}{6} \log (2)+\int _0^1\int _0^1\frac{\log \left(1-\frac{x y}{2}\right)}{1-x y}dxdy$$ and try to integrate (See for example https://arxiv.org/abs/math/0502034 [Thirty-Two Goldbach Variations by Borwein and Bradley] section 5.3, for the change of variables integration technique they use to see if it works with this variation) – James Arathoon May 23 '20 at 23:49
  • Thanks so much! That’s very interesting. – MichaelCatliMath May 24 '20 at 02:41
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    You may have a look at this question: https://math.stackexchange.com/questions/275643/proving-an-alternating-euler-sum-sum-k-1-infty-frac-1k1-h-kk (especially robjohn's answer to one of the bonus questions). –  May 26 '20 at 12:38

1 Answers1

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The closed form follows on plugging $x=1/2$ in the generating function

$$\small{\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{n}x^{n}=\operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\ln(1-x)\operatorname{Li}_2(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)}$$

note that $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^2(2)$ and $\operatorname{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$

Different approach

Recall the generalization

$$\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}dx=(-1)^a a! \sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n}$$

With $a=1$ we have

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n2^n}=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx$$ which is trivial.

Ali Shadhar
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