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$\ds{}$
Tenga en cuenta que
\begin{align}
H_{n}&=\int_{0}^{1}{1 - t^{n} \over 1 - t}\,\dd t
=-n\int_{0}^{1}\ln\pars{1 - t}t^{n - 1}\,\dd t
\end{align}
donde integramos por partes.
tal que
\begin{align}
\color{#66f}{\large\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{H_{n} \over n}}&
=-\int_{0}^{1}\ln\pars{1 - t}\sum_{n = 1}^{\infty}\pars{-t}^{n - 1}\,\dd t
=-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t
\\[3mm]&=-\int_{0}^{1}{\ln\pars{t} \over 2 - t}\,\dd t
=-\int_{0}^{1/2}{\ln\pars{2t} \over 1 - t}\,\dd t
=-\int_{0}^{1/2}{\ln\pars{1 - t} \over t}\,\dd t
\\[3mm]&=\int_{0}^{1/2}{{\rm Li}_{1}\pars{t} \over t}\,\dd t
=\int_{0}^{1/2}{\rm Li}_{2}'\pars{t}\,\dd t={\rm Li}_{2}\pars{\half}
=\\[3mm]&=\color{#66f}{\large{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}
\approx 0.5822
\end{align}
$\large\mbox{See this link}$.
