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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.

$$ \sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\dfrac{1}{i+1} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\int_{0}^{1}x^{i}dx = ? $$

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  • $\begingroup$ interestingly enough, $\sum_{n = 1}^{\infty}\frac{1}{2^nn^2}$ is equal to the same value. I wonder if there's a simple way to prove the two are equal. $\endgroup$ – recursive recursion Jul 12 '14 at 21:33
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    $\begingroup$ @recursiverecursion That serie is $\displaystyle{\large{\rm Li}_{\,\,2}\left(1 \over 2\right)}$. See my answer below. $\endgroup$ – Felix Marin Jul 12 '14 at 23:07
  • $\begingroup$ Check this. $\endgroup$ – Mhenni Benghorbal Jul 13 '14 at 2:14
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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$$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} \!= -\! \int_{-1}^{0}\dfrac{\ln(1-x)}{x(1-x)} dx = -\!\int_{-1}^{0}\left(\dfrac{\ln(1-x)}{x}\! + \!\dfrac{\ln(1-x)}{1-x}\right) \! dx=\dfrac{\pi^2}{12} - \dfrac{1}{2}\ln^2 2.$$

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

$\large\mbox{See this link}$.

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  • $\begingroup$ +1. BTW, I think the OP relates to $$\sum_{n=1}^\infty\frac{1}{2^n\ n^2}.$$ $\endgroup$ – Tunk-Fey Jul 14 '14 at 19:14
  • $\begingroup$ @Tunk-Fey That was the coincidence which $\color{#88f}{\tt\mbox{@recursive recursion}}$ already commented above. Thanks. $\endgroup$ – Felix Marin Jul 14 '14 at 20:15

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