# Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^a H_n}{2^n}$

Is there a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^a H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$

This is a generalization of my previous question that was just a special case for $a=4$.

• What this tells me is that you are unable to evaluate the inner sum of your previous example. Perhaps that's what you really should be asking about, – Ron Gordon May 8 '13 at 2:38
• Is $a$ a positive integer? – Mhenni Benghorbal Aug 14 '13 at 19:11

## 1 Answer

Replace $n^a$ by $$n^a=\frac{1}{\Gamma(-a)}\int_0^{\infty}t^{-a-1}e^{-nt}dt,$$ and also use the trick decribed here to transform your sum into \begin{align}\frac{1}{\Gamma(-a)}\int_0^{\infty}t^{-a-1}\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\frac{1}{k}\left(-\frac{e^{-t}}{2}\right)^n dt &=\frac{1}{\Gamma(-a)}\int_0^{\infty}\frac{t^{-a-1}}{1+\frac12 e^{-t}}\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{e^{-t}}{2}\right)^k dt=\\ &=\frac{-1}{\Gamma(-a)}\int_0^{\infty}\frac{t^{-a-1}}{1+\frac12 e^{-t}}\ln\left(1+\frac12 e^{-t}\right) dt.\end{align} I dont't think, however, this can be simplified further. One can compute this integral for $a$ given by negative integers and the answer should be given by polylogarithms of increasing order. I can hardly imagine a nice function that would interpolate such values.

In fact, for negative integer $a$ one can write an explicit general formula for the sum in the form $$\sum_{n=1}^{\infty}\frac{H_n}{n^{N-1}}x^n=\gamma\, \mathrm{Li}_{N-1}(x)+\left[\frac{\partial}{\partial s}\left\{x\Gamma(1+s)\cdot {}_{N+1}F_{N}\left[\begin{array}{c}1,\ldots,1,1+s\\ 2,\ldots,2\end{array};x\right]\right\}\right]_{s=1},$$ which follows from the series representation for $_pF_q$ and your last formula $H_n=\gamma+\psi(n+1)$.