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Here is a challenging double series question I wanna share with you

$$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{H_n}{kn (k+n)^3}$$

together with the question: what tools would you like employ for computing it?

EDIT: thank you, no need for further work. I know how to do it now, but it's a long way.

Martin Sleziak
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  • $\approx 0.22202673650444240323022298994402744342722603057060149438381566111913% 70686798303720960651610066092228$. – Felix Marin Nov 15 '14 at 21:48
  • Have you tried writing a similar double sum, with $H_k$ instead of $H_n$, adding them together, and then exploiting the symmetry ? – Lucian Nov 16 '14 at 08:09
  • @FelixMarin That agrees to 26 decimal places with r9m's value. My guess is that is the limit of the precision of your computation. – robjohn Nov 16 '14 at 23:47

1 Answers1

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$$\begin{align}&\sum\limits_{n,k=1}^{\infty} \dfrac{H_n}{nk(n+k)^3} \\&= \sum\limits_{m=2}^{\infty}\frac{1}{m^3} \sum\limits_{n=1}^{m-1}\frac{H_n}{n(m-n)} \\&= \sum\limits_{m=2}^{\infty}\frac{1}{m^4} \left(\sum\limits_{n=1}^{m-1}\frac{H_n}{n}+\frac{H_n}{m-n}\right) \\ &=\sum\limits_{m=2}^{\infty}\frac{1}{m^4} \left(\frac{1}{2}(H_{m-1}^2 + H_{m-1}^{(2)})+ H_{m}^2 - H_m^{(2)}\right) \\ & =\sum\limits_{m=1}^{\infty}\frac{1}{m^4} \left(\frac{3}{2}H_{m}^2 - \frac{H_m}{m} -\frac{1}{2} H_m^{(2)}\right)\end{align}$$

where, we used: $\displaystyle 2\sum\limits_{n=1}^{m}\frac{H_n}{n} = H_m^2 + H_m^{(2)}$ and $\displaystyle \sum\limits_{n=1}^{m-1}\frac{H_n}{m-n} = H_m^2 - H_m^{(2)}$

These series can be computed with elementary tools. Please check here.The closed form is $\frac{215}{48}\zeta(6) - 3\zeta^2(3)$.

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