Evaluate the following series sum.

Question

Problem

I’m trying to evaluate the following series sum

\begin{equation} S_{j}(z) = \sum_{k=1}^{\infty} \frac{2 H_{k} z^{k+2}}{(k+1)(k+2)^{j}} \end{equation}

Where

\begin{equation} H_{k} = \sum_{n=1}^{k} \frac{1}{n} \end{equation} is the $k^{\text{th}}$ Harmonic Number.

Notation

$k, j$ and $n$ are all positive definite Integers.

$z$ is a Real Number.

First Step

The first step that I did was to expand the denominator as a Partial Fraction sum

\begin{equation} \frac{1}{(k+1)(k+2)^{j}} = \frac{1}{k+1} - \sum_{m = 1}^{j}\frac{1}{(k+2)^{m}} \end{equation} Which then gives

\begin{equation} S_{j}(z) = \sum_{k=1}^{\infty} \frac{2 H_{k} z^{k+2}}{(k+1)} - \sum_{m=1}^{j} \sum_{k=1}^{\infty} \frac{2 H_{k} z^{k+2}}{(k+2)^{m}} \end{equation} Here the first sum can be written in terms of a known result \begin{equation} (\ln(1-z))^{2} = \sum_{k=1}^{\infty} \frac{2 H_{k} z^{k+1}}{k+1} \end{equation} and so we have \begin{equation} S_{j}(z) =z (\ln(1-z))^{2} - \sum_{m=1}^{j} \sum_{k=1}^{\infty} \frac{2 H_{k} z^{k+2}}{(k+2)^{m}} \end{equation} At this point I got stuck and couldn’t proceed any further. Any assistance here would be greatly appreciated.

Answer 1

This not an answer.

$$S_{j}(z) = \sum_{k=1}^{\infty} \frac{2 H_{k} }{(k+1)(k+2)^{j}}\,z^{k+2}$$

$$S_1(z)=-2 z+\log ^2(1-z)+2 (z-1) \log (1-z)$$ $$3S_2(z)=-6 \text{Li}_3(1-z)+6 \text{Li}_2(1-z) (\log (1-z)-1)-18 z+$$ $$3 \log (1-z) (4 z-2 \log (z)+\log (1-z) (\log (z)+1)-4)+6 \zeta (3)+\pi ^2$$ I have been unable (using a CAS) to compute any $S_j(z)$ for any $j>2$ but we can expect aa bunch of polylogarithms and zeta functions.

For the case where $z=1$, there is no problem.