TITLE: Evaluating $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$
QUESTION [23 upvotes]: I would appreciate to understand the main steps giving the evaluation of this series: $$ S=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$$ where $H_n$ is the harmonic number. I've tried with no success to obtain this sum with the help of Wolfram Alpha.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{S \equiv \sum_{n\ =\ 1}^{\infty}\pars{-1}^{n - 1}\,{H_{2n} \over n}}$

With the Harmonic Number Generating Function
  $$
-\,{\ln\pars{1 - x} \over 1 - x} = \sum_{n\ =\ 1}^{\infty}H_{n}x^{n} 
$$

we'll have:
\begin{align}
-\,{\ln\pars{1 - x} \over 1 - x} &= \sum_{n\ =\ 1}^{\infty}H_{n}x^{n}
\\
-\,{\ln\pars{1 + x} \over 1 + x} &= \sum_{n\ =\ 1}^{\infty}H_{n}\pars{-x}^{n}
\\[5mm]
-\,{\ln\pars{1 - x} \over 1 - x} -{\ln\pars{1 + x} \over 1 + x}
&= 2\sum_{n\ =\ 1 \atop n\ \mbox{even}}^{\infty}H_{n}x^{n}
=2\sum_{n\ =\ 1}^{\infty}H_{2n}x^{2n}
\\[5mm]
-\int_{0}^{x}\bracks{{\ln\pars{1 - t} \over 1 - t}
+{\ln\pars{1 + t} \over 1 + t}}\,{\dd t \over t}
&=\sum_{n\ =\ 1}^{\infty}H_{2n}\,{x^{2n} \over n}
\end{align}

We'll set $\ds{x = \ic}$ such that
  \begin{align}
S&=\color{#66f}{\large\sum_{n\ =\ 1}^{\infty}\pars{-1}^{n - 1}\,{H_{2n} \over n}}
=\int_{0}^{\ic}\bracks{{\ln\pars{1 - t} \over 1 - t}
+{\ln\pars{1 + t} \over 1 + t}}\,{\dd t \over t}
\\[5mm]&=\sum_{\sigma\ =\ \pm}\int_{0}^{\ic}
{\ln\pars{1 - \sigma t} \over \pars{1 - \sigma t}t}\,\dd t
=\sum_{\sigma\ =\ \pm}\bracks{\sigma\int_{0}^{\ic}
{\ln\pars{1 - \sigma t} \over 1 - \sigma t}\,\dd t
+\int_{0}^{\ic}{\ln\pars{1 - \sigma t} \over t}\,\dd t}
\\[5mm]&=\sum_{\sigma\ =\ \pm}\bracks{-\,\half\,\left.\ln^{2}\pars{1 - \sigma t}
\right\vert_{t\ =\ 0}^{t\ =\ \ic}
+\int_{0}^{\sigma\ic}{\ln\pars{1 - t} \over t}\,\dd t}
\\[5mm]&=\sum_{\sigma\ =\ \pm}\bracks{-\,\half\,\ln^{2}\pars{1 - \sigma\ic}
-\int_{0}^{\sigma\ic}{\rm Li}_{2}'\pars{t}\,\dd t}
\\[5mm]&=-\,\Re\bracks{\ln^{2}\pars{1 - \ic} + 2\,{\rm Li}_{2}\pars{\ic}}
=\color{#66f}{\large -\,\Re\ln^{2}\pars{1 - \ic} - 2\,\Re{\rm Li}_{2}\pars{\ic}}
\qquad\pars{1}
\end{align}

However,
\begin{align}
\color{#00f}{\Re\ln^{2}\pars{1 - \ic}}
&=\Re\braces{\bracks{\half\,\ln\pars{2} - {\pi \over 4}\,\ic}^{2}}
=\color{#00f}{{1 \over 4}\,\ln^{2}\pars{2} - {\pi^{2} \over 16}}
\\[1cm]
\color{#00f}{\Re{\rm Li}_{2}\pars{\ic}}&
=\Re\sum_{n\ =\ 1}^{\infty}{\ic^{n} \over n^{2}}
=\sum_{n\ =\ 1}^{\infty}{\ic^{2n} \over \pars{2n}^{2}}
={1 \over 4}\sum_{n\ =\ 1}^{\infty}{\pars{-1}^{n} \over n^{2}}
\\[5mm]&={1 \over 4}\bracks{-\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2}}
+\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}}
\\[5mm]&={1 \over 4}\bracks{-\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}
+ \sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}
+\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n}^{2}}}
=-\,{1 \over 8}\ \underbrace{\sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}}
_{\ds{\color{#c00000}{\pi^{2} \over 6}}}\ =\ \color{#00f}{-\,{\pi^{2} \over 48}}
\end{align}

With the previous partial result ( see expression $\pars{1}$ ) and these ones:
  \begin{align}
S &\equiv \color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}\pars{-1}^{n - 1}\,{H_{2n} \over n}}
=-\bracks{{1 \over 4}\,\ln^{2}\pars{2} - {\pi^{2} \over 16}}
- 2\pars{-\,{\pi^{2} \over 48}}
\\[5mm]&=\color{#66f}{\large{5\pi^{2} \over 48} - {1 \over 4}\,\ln^{2}\pars{2}}
\approx {\tt 0.9080}
\end{align}