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

\begin{align}&\color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k} \int_{0}^{1}{1 - t^{k} \over 1 - t}\,\dd t \\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} \ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t \\[3mm]&=-\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 =-\,\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 \end{align} where $\ds{{\rm Li}_{s}\pars{z}}$ is a PolyLogarithm Function and we'll use well known properties of them as explained in the above mentioned link.

$$ \color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} ={\rm Li}_{2}\pars{\half} $$

\begin{align}&\color{#66f}{\large% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} ={\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} =\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}} \end{align}

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

\begin{align}&\color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k} \int_{0}^{1}{1 - t^{k} \over 1 - t}\,\dd t \\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} \ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t \\[3mm]&=-\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 =-\,\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 \end{align} where $\ds{{\rm Li}_{s}\pars{z}}$ is a PolyLogarithm Function and we'll use well known properties of them as explained in the above mentioned link.

Then, $$ \color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\int_{0}^{1/2}{\rm Li}_{2}'\pars{t}\,\dd t ={\rm Li}_{2}\pars{\half} - {\rm Li}_{2}\pars{0} =\color{#c00000}{{\rm Li}_{2}\pars{\half}} $$

$\ds{{\rm Li}_{2}\pars{\half}}$ is given in the above mentioned link: \begin{align}&\color{#66f}{\large% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} ={\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} =\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}} \end{align}

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

\begin{align}&\color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k} \int_{0}^{1}{1 - t^{k} \over 1 - t}\,\dd t \\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} \ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t \\[3mm]&=-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t =-\,\half\int_{0}^{1}{\ln\pars{t} \over 1 - t/2}\,\dd t =-\,\half\sum_{n = 0}^{\infty}\int_{0}^{1}\ln\pars{t}\pars{t \over 2}^{n}\,\dd t \end{align}

$$ \int_{0}^{1}\ln\pars{t}t^{n}\,\dd t =\lim_{\mu \to n}\partiald{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t =-\,{1 \over \pars{n + 1}^{2}} $$

\begin{align}&\color{#66f}{\large% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}} ={\rm Li}_{2}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} \\[3mm]&=\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}} \end{align}

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

\begin{align}&\color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k} \int_{0}^{1}{1 - t^{k} \over 1 - t}\,\dd t \\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} \ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t \\[3mm]&=-\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 =-\,\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 \end{align} where $\ds{{\rm Li}_{s}\pars{z}}$ is a PolyLogarithm Function and we'll use well known properties of them as explained in the above mentioned link.

$$ \color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} ={\rm Li}_{2}\pars{\half} $$

\begin{align}&\color{#66f}{\large% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} ={\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} =\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}} \end{align}

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

\begin{align}&\color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\bracks{\Psi\pars{k + 1} - \Psi\pars{1}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} {1 - t^{k} \over 1 - t}\,\dd t \\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} \ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t \\[3mm]&=-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t =-\,\half\int_{0}^{1}{\ln\pars{t} \over 1 - t/2}\,\dd t =-\,\half\sum_{n = 0}^{\infty}\int_{0}^{1}\ln\pars{t}\pars{t \over 2}^{n}\,\dd t \end{align}

$$ \int_{0}^{1}\ln\pars{t}t^{n}\,\dd t =\lim_{\mu \to n}\partiald{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t =-\,{1 \over \pars{n + 1}^{2}} $$

\begin{align}&\color{#66f}{\large% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}} ={\rm Li}_{2}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} \\[3mm]&=\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}} \end{align}

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

\begin{align}&\color{#c00000}{% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k} \int_{0}^{1}{1 - t^{k} \over 1 - t}\,\dd t \\[3mm]&=\sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over k}\int_{0}^{1} \ln\pars{1 - t}\pars{-kt^{k - 1}}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\sum_{k = 1}^{\infty}\pars{-t}^{k - 1}\,\dd t \\[3mm]&=-\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t =-\,\half\int_{0}^{1}{\ln\pars{t} \over 1 - t/2}\,\dd t =-\,\half\sum_{n = 0}^{\infty}\int_{0}^{1}\ln\pars{t}\pars{t \over 2}^{n}\,\dd t \end{align}

$$ \int_{0}^{1}\ln\pars{t}t^{n}\,\dd t =\lim_{\mu \to n}\partiald{}{\mu}\int_{0}^{1}t^{\mu}\,\dd t =-\,{1 \over \pars{n + 1}^{2}} $$

\begin{align}&\color{#66f}{\large% \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} H_{\rm k}\over k}} =\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}} ={\rm Li}_{2}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} \\[3mm]&=\color{#66f}{\large\half\bracks{\zeta\pars{2} - \ln^{2}\pars{2}}} \end{align}