# Alternating Euler sums

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

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

• Recall that $H_n$ are harmonic numbers. – Jean Marie Aug 5 '16 at 6:40
• Have a look at (math.stackexchange.com/q/457371) – Jean Marie Aug 5 '16 at 6:43
• I have the p=3 case. Lookibg forward to the 4 th power case. The integrations are quite complex. – Aditya Narayan Sharma Aug 5 '16 at 7:00
• 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. – Jean Marie Aug 5 '16 at 7:05

In Euler Sums and Contour Integral Representations by P. Flajolet and B. Salvy they state in Theorem 7.1 (Sitaramachandra Rao) for odd weight $q+1$:
We obtain for odd $q+1=5$:
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.