Prove that ${\sum\limits_{n=1}^{\infty}}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac12\ln^2 2$

You may consider the standard identity $$\sum_{n=1}^{\infty}H_n x^{n-1} = -\dfrac{\ln(1-x)}{x(1-x)} \quad -1 < x<1,\,x\neq0.$$ Then integrate from $x=-1$ to $x=0$ to obtain easily$$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} \!= -\! \int_{-1}^{0}\dfrac{\ln(1-x)}{x(1-x)} dx = -\!\int_{-1}^{0}\left(\dfrac{\ln(1-x)}{x}\! + \!\dfrac{\ln(1-x)}{1-x}\right) \! dx=\dfrac{\pi^2}{12} - \dfrac{1}{2}\ln^2 2.$$

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

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