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$$\displaystyle \sum_{m=0}^{\infty}{\frac{{\left({H}_{m}^{(1)}\right)}^{2} - {H}_{m}^{(2)}}{{(m+1)}^{6}}}$$

where $ \displaystyle {{H}_{k}}^{(r)} = \sum_{i=1}^{k}{\frac{1}{{i}^{r}}} $

I have no idea how to approach to this problem. I want a full closed form and an idea about how to approach these types of problems(use contour integration? I don't know, I haven't tried).

Edit: I would like to have the answer in terms of zeta function.

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  • $\begingroup$ One method. Compute many decimals, then put them in ISC to see if they are recognized. isc.carma.newcastle.edu.au $\endgroup$ – GEdgar Jul 19 '15 at 12:47
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Recalling the definition of Euler's sums $$s_{h}\left(m,n\right)=\sum_{k\geq1}\frac{H_{k}^{m}}{\left(k+1\right)^{n}} $$ $$\sigma_{h}\left(m,n\right)=\sum_{k\geq1}\frac{H_{k}^{\left(m\right)}}{\left(k+1\right)^{n}} $$ and using the fact that $H_{0}=0 $ $$\sum_{m\geq0}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}=\sum_{m\geq1}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}=s_{h}\left(2,6\right)-\sigma_{h}\left(2,6\right) $$ and they can be write in terms of not so nice combinations of zeta function (see for example here and here). Using the identity $22$ here, we have explicitly $$\sum_{m\geq0}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}=s_{h}\left(2,6\right)-\sigma_{h}\left(2,6\right)=14\zeta\left(8\right)+\zeta\left(2\right)\zeta\left(6\right)-3\sum_{k=0}^{4}\zeta\left(6-k\right)\zeta\left(k+2\right)+ $$ $$+\frac{1}{3}\sum_{k=2}^{4}\zeta\left(6-k\right)\sum_{j=1}^{k-1}\zeta\left(j+1\right)\zeta\left(k+1-j\right). $$

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  • $\begingroup$ Thanks, that was helpful. But, it would be nicer of you if you provide with the "not so nice combinations of zeta function". $\endgroup$ – Kartik Sharma Jul 20 '15 at 13:06
  • $\begingroup$ @KartikSharma I wrote the result. $\endgroup$ – Marco Cantarini Jul 20 '15 at 13:39
  • $\begingroup$ Oh, thanks! Actually, I wanted to find the following - $ \displaystyle \frac{1}{120} \int_{0}^{\infty}{\frac{{log}^{2}(1-{e}^{-x}){x}^{5}}{{e}^{x}-1} dx} $ and I believe that they both have the same values. So, is it true? $\endgroup$ – Kartik Sharma Jul 20 '15 at 13:42
  • $\begingroup$ Can you please help with this? I am asking so because the answer of the above integral was a very simple zeta function combination(I don't know the exact answer, just that it was a simple combination), so I doubt what I have done. $\endgroup$ – Kartik Sharma Jul 20 '15 at 13:46
  • $\begingroup$ @KartikSharma The series I get from the integral is $$\sum_{n\geq1}H_{n}\sum_{k\geq1}\frac{1}{k\left(k+n\right)^{6}}$$ (hoping there is no mistakes :D ) which can be rewritten as $$\sum_{m=2}^{\infty}\frac{1}{m^{6}}\sum_{n=1}^{m-1}\frac{H_{n}}{m-n}=\sum_{m=1}^{\infty}\frac{H_{m}^{2}-H_{m}^{\left(2\right)}}{\left(m+1\right)^{6}}$$ so I think your calculation are right. $\endgroup$ – Marco Cantarini Jul 20 '15 at 18:18

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