TITLE: Evaluate:: $ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 +\cdots + \frac 1n\right) $
QUESTION [34 upvotes]: How to evaluate the series:
$$  2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$
According to Mathematica, this converges to $ (\log 2)^2 $.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&2\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n + 1}
\pars{1 + {1 \over 2} + \cdots + {1 \over n}} =
-2\sum_{n = 1}^{\infty}\pars{-1}^{n}\, H_{n}\int_{0}^{1}x^{n}\,\dd x =
-2\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\pars{-x}^{n}\,\dd x
\\[5mm] = &\
-2\int_{0}^{1}\braces{-\,{\ln\pars{1 - \bracks{-x}} \over 1 - \pars{-x}}}\,\dd x =
2\int_{0}^{1}{\ln\pars{1 +x} \over 1 + x}\,\dd x =
\left.\ln^{2}\pars{1 + x}\,\right\vert_{\ x\ =\ 0}^{\ x\ =\ 1} =
\bbx{\ds{\ln^{2}\pars{2}}}
\end{align}