# 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, May 8, 2013 at 2:38
• Is $a$ a positive integer? Aug 14, 2013 at 19:11

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 $H_n=\gamma+\psi(n+1)$.