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Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Evaluate the following sum

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1}$$

Here $\mathcal{H}_n^2$ is the square harmonic number, e.g $\left ( \sum \limits_{k=1}^{n} \frac{1}{k} \right )^2$. I don't know how to tackle this. One idea of mine was the following:

\begin{align*} \mathcal{S} &=\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n^2 \int_{0}^{1}x^n \, {\rm d}x \\ &= \int_{0}^{1}\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n^2 x^n \, {\rm d}x \end{align*}

and the last sum evaluates to ?

Motivation: The series comes from this question . I tried a different approach. This is what I tried:

\begin{align*} \int_{0}^{1} \frac{\log(1+x) \log(1-x)}{1+x} \, {\rm d}x&= \int_{0}^{1}\log(1-x) \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n x^n \, {\rm d}x\\ &=\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \int_{0}^{1}x^n \log(1-x) \, {\rm d}x \\ &= -\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \cdot \frac{\mathcal{H}_{n+1}}{n+1} \\ &= -\sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \left ( \frac{\mathcal{H}_n + \frac{1}{n+1}}{n+1} \right )\\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} \end{align*}

The right hand Euler sum appears to be elementary. I have not evaluated it though. But now we known that:

$$\bbox[blue, 2pt]{\color{white}{-\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} = \frac{1}{3}\log^3 2-\frac{\pi^2}{12}\log 2+\frac{\zeta(3)}{8}}}$$

from the linked question. So, I expect both of these series to have a closed form. Any help how to proceed with the first series? If anyone wishes to give a go for the second (right hand Euler sum) be my guest.

Addendum: For the second Euler sum we have that:

\begin{align*} -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n}{\left ( n+1 \right )^2} &= \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n \int_{0}^{1}x^n \log x \, {\rm d}x\\ &=\int_{0}^{1}\log x \sum_{n=1}^{\infty} (-1)^{n-1} \mathcal{H}_n x^n \, {\rm d}x \\ &= \int_{0}^{1}\frac{\log (1+x) \log x}{1+x}\, {\rm d}x\\ &=-\frac{\zeta(3)}{8} \end{align*}

The integral is elementary. Thus, it appears that the sum I seek evaluates to

$$\frac{1}{3} \log^3 2 - \frac{\pi^2}{12} \log 2 + \frac{\zeta(3)}{8} = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1} - \frac{\zeta(3)}{8} \Rightarrow \\\\\\ \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n^2}{n+1}= -\frac{1}{3} \log^3 2 +\frac{\pi^2}{12} \log 2 - \frac{\zeta(3)}{4}$$

Of course this is an indirect way of evaluating the sum. Any other way of tackling it?

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  • $\begingroup$ Added some more details, namely the derivation of the second sum. $\endgroup$ – Tolaso Jun 5 '16 at 13:15
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Some preliminary lemma.
Lemma 1. $$ \sum_{n\geq 1} H_n x^n = \frac{\log(1-x)}{1-x}. $$ Lemma 2. By Lemma $1$, $$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n+1} = \frac{1}{2}\log^2(1-x),\qquad \sum_{n\geq 1}\frac{H_n+H_{n+1}}{n+1}x^{n}=\frac{-x+\log^2(1-x)+\text{Li}_2(x)}{x}.$$ Lemma 3. Since $H_{n+1}^2-H_n^2 = \frac{H_n+H_{n+1}}{n+1}$, $$ \sum_{n\geq 1}H_{n}^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}.$$ Lemma 4. By Lemma 3, $$ \sum_{n\geq 1}\frac{(-1)^{n+1} H_{n}^2}{n+1} = -\int_{0}^{1}\frac{\log^2(1+x)+\text{Li}_2(-x)}{1+x}\,dx=-\frac{\log^3(2)}{3}-\color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}.$$ The problem boils down to the evaluation of the last integral. By integration by parts, it is: $$ \color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}=-\frac{\pi^2}{12}\log(2)+\color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx}\tag{1} $$ but: $$ \color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx} = -2\int_{0}^{1}\frac{\log(1+x)\log(x)}{1+x}\,dx=\color{blue}{\frac{\zeta(3)}{4}}\tag{2}$$ and the proof is complete.

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  • $\begingroup$ Jack thank you. But it appears that we have an unmatch. I havew $\frac{\pi^2}{12} \log 2$ while you have $-$. Anyway (+1) for your elementary solution. $\endgroup$ – Tolaso Jun 7 '16 at 7:28
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

$\underline{\mbox{Just one of the integrals you didn't evaluate}\ }$ :

With $\ds{0 < \mu < 1}$: \begin{align} &\color{#f00}{% \int_{0}^{1 - \mu}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1 + x}\,\dd x} = \half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} + \half\,\int_{0}^{1 - \mu}{\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x \\[3mm] = &\ \half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} + \half\,\int_{1}^{2 - \mu}{\ln^{2}\pars{x} \over 2 - x}\,\dd x \\[3mm] = &\ \half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} + \half\,\int_{1/2}^{1 - \mu/2}{\ln^{2}\pars{2x} \over 1 - x}\,\dd x \\[3mm] = &\ \half\,\ln^{2}\pars{2 - \mu}\ln\pars{\mu} - \half\ln\pars{{\mu \over 2}}\ln^{2}\pars{2 - \mu} + \int_{1/2}^{1 - \mu/2}{\ln\pars{1 - x} \over x}\,\ln\pars{2x}\,\dd x \end{align}


When $\ds{\mu \to 0^{+}}$: \begin{align} &\color{#f00}{% \int_{0}^{1}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1 + x}\,\dd x} = \half\,\ln^{3}\pars{2} -\int_{1/2}^{1}\mathrm{Li}_{2}'\pars{x}\ln\pars{2x} \,\dd x \\[3mm] = &\ \half\,\ln^{3}\pars{2} - \mathrm{Li}_{2}\pars{1}\ln\pars{2} + \int_{1/2}^{1}\mathrm{Li}_{3}'\pars{x}\,\dd x \\[3mm] = &\ \color{#f00}{\half\,\ln^{3}\pars{2} - \mathrm{Li}_{2}\pars{1}\ln\pars{2} + \mathrm{Li}_{3}\pars{1} - \mathrm{Li}_{3}\pars{\half}} \end{align}
$$ \mbox{With}\quad \left\lbrace\begin{array}{rcl} \ds{\mathrm{Li}_{2}\pars{1}} & \ds{=} & \ds{{\pi^{2} \over 6}} \\[2mm] \ds{\mathrm{Li}_{3}\pars{1}} & \ds{=} & \ds{\zeta\pars{3}} \\[2mm] \ds{\mathrm{Li}_{3}\pars{\half}} & \ds{=} & \ds{{1 \over 24}\bracks{21\zeta\pars{3} + 4\ln^{3}\pars{2} - 2\pi^{2}\ln\pars{2}}} \end{array}\right. $$

we'll get $$ \color{#f00}{% \int_{0}^{1}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1 + x}\,\dd x} = \color{#f00}{{1 \over 24}\bracks{3\zeta\pars{3} + 8\ln^{3}\pars{2} -2\pi^{2}\ln\pars{2}}} \approx -0.3028 $$

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