TITLE: Prove that ${\sum\limits_{n=1}^{\infty}}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac12\ln^2 2$
QUESTION [16 upvotes]: We know that $H_n = \sum_{j=1}^{n}{1 \over j}$. Article in The Sum of Certain Series Related To Harmonic Numbers of Omran Kolba, we have proof of this identity which involves some advanced concepts. 
I tried to turn the sum into a definite integral and could not. I appreciate any help.
$$
\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\dfrac{1}{i+1} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\int_{0}^{1}x^{i}dx = ?
$$

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
Note that
\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}
where we integrated by parts.

Then,
\begin{align}
&\bbox[10px,#ffd]{\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
\\[5mm] = &\
-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t =
-\int_{0}^{1}{\ln\pars{t} \over 2 - t}\,\dd t
=-\int_{0}^{1/2}{\ln\pars{2t} \over 1 - t}\,\dd t
\\[5mm] = &\
-\int_{0}^{1/2}{\ln\pars{1 - t} \over t}\,\dd t =
\int_{0}^{1/2}{{\rm Li}_{1}\pars{t} \over t}\,\dd t
=\int_{0}^{1/2}{\rm Li}_{2}'\pars{t}\,\dd t
\\[5mm] = &\ {\rm Li}_{2}\pars{\half} =
\bbox[10px,border:1px groove navy]
{{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}
\approx 0.5822
\end{align}
$\large\mbox{See this link}$.