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$ \displaystyle \sum_{n=1}^{\infty} (-1)^{(n-1)} \frac{{\rm H}_n}{n^p} $

Does this have a nice closed form? I am trying to evaluate the case of p=4 also

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  • $\begingroup$ Recall that $H_n$ are harmonic numbers. $\endgroup$ – Jean Marie Aug 5 '16 at 6:40
  • $\begingroup$ Have a look at (math.stackexchange.com/q/457371) $\endgroup$ – Jean Marie Aug 5 '16 at 6:43
  • $\begingroup$ I have the p=3 case. Lookibg forward to the 4 th power case. The integrations are quite complex. $\endgroup$ – Aditya Narayan Sharma Aug 5 '16 at 7:00
  • $\begingroup$ I advise you, for another time, to make a prior web search and indicate through the relevant site(s) what is the degree of knowledge you have about the issue at hand. $\endgroup$ – Jean Marie Aug 5 '16 at 7:05
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In Euler Sums and Contour Integral Representations by P. Flajolet and B. Salvy they state in Theorem 7.1 (Sitaramachandra Rao) for odd weight $q+1$:

\begin{align*} \sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}{n^q} &=\frac{q+1}{2}\left(1-2^{-q}\right)\zeta(q+1)-\frac{1}{2}\zeta(q+1)\\ &\qquad-\sum_{k=1}^{q/2-1}\left(1-2^{1-2k}\right)\zeta(2k)\zeta(q+1-2k) \end{align*}

We obtain for odd $q+1=5$:

\begin{align*} \sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}{n^4} &=\frac{5}{2}\left(1-2^{-4}\right)\zeta(5)-\frac{1}{2}\zeta(5)\ -\left(1-2^{-1}\right)\zeta(2)\zeta(3)\\ &=\frac{59}{32}\zeta(5)-\frac{\pi^2}{12}\zeta(3) \end{align*}

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A very simple solution to the case $p=4$ that can be extended to all values, $p=2m, \ m\ge1$, may be found in https://math.stackexchange.com/q/3269815.

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