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.


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  • $\begingroup$ Do you have a conjecture for the value? Where did this show up? Mathematica is taking a particularly long time evaluating this ... $\endgroup$ – Sandeep Silwal Dec 6 '18 at 22:38
  • 1
    $\begingroup$ 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 $\frac{17}{4}\zeta(4).$ $\endgroup$ – Isak Dec 6 '18 at 22:50
  • $\begingroup$ The answer is $$\frac{\pi^4}{32}$$ $\endgroup$ – user178256 Dec 6 '18 at 22:54
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    $\begingroup$ How can this be proven? $\endgroup$ – Isak Dec 6 '18 at 23:19

There's a nasty trick here:

$$ \frac{H_{2n}-\frac{1}{2}H_n}{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{2k-1}=\int_{0}^{1}x^{2n}\cdot \log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}.\tag{1} $$ 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, $\color{red}{\frac{\pi^4}{32}}$.

  • $\begingroup$ nasty, may be, but beautiful for sure ! $\endgroup$ – Claude Leibovici Dec 7 '18 at 5:45
  • $\begingroup$ Thank You very much! $\endgroup$ – Isak Dec 7 '18 at 10:22

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