Alternating Euler sums

Question

$ \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

Answer 1

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*}