Evaluation of Euler - type sum $\sum\limits_{n=1}^{\infty}\frac{\left(H_{2n}-\frac{1}{2}H_{n}\right)^{2}}{n^{2}}$ [closed]

How can one evaluate the sum $$\displaystyle\sum_{n=1}^{\infty}\frac{\left(H_{2n}-\frac{1}{2}H_{n}\right)^{2}}{n^{2}}$$? Here $$H_{n}$$ denotes the $$n$$-th harmonic number.

closed as off-topic by Masacroso, Clarinetist, T. Bongers, user10354138, KReiserDec 7 '18 at 8:20

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• Do you have a conjecture for the value? Where did this show up? Mathematica is taking a particularly long time evaluating this ... – Sandeep Silwal Dec 6 '18 at 22:38
• Unfortunately, I have no conjecture. I'm curious of the value of this sum. The similar sum $\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{2}}$ is known to be $\frac{17}{4}\zeta(4).$ – Isak Dec 6 '18 at 22:50
• The answer is $$\frac{\pi^4}{32}$$ – user178256 Dec 6 '18 at 22:54
• How can this be proven? – Isak Dec 6 '18 at 23:19

$$\frac{H_{2n}-\frac{1}{2}H_n}{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{2k-1}=\int_{0}^{1}x^{2n}\cdot \log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}.\tag{1}$$ In particular $$\sum_{n\geq 1}\left(\frac{H_{2n}-\frac{1}{2}H_n}{n}\right)^2=\iint_{(0,1)^2}\frac{x^2 y^2}{1-x^2 y^2}\log\left(\frac{1+x}{1-x}\right)\log\left(\frac{1+y}{1-y}\right)\frac{dx\,dy}{xy}.\tag{2}$$ Since $$x\mapsto\frac{1-x}{1+x}$$ is an involution the last integral equals $$\iint_{(0,1)^2}\frac{(1-x)(1-y)\log(x)\log(y)}{(1+x)(1+y)(x+y)(1+xy)}\,dx\,dy \tag{3}$$ which is fairly simple to compute via Maclaurin series. It is, indeed, $$\color{red}{\frac{\pi^4}{32}}$$.