# Evaluation of Euler-type sum $\displaystyle\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{(2n+1)^{2}}$ [closed]

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

## closed as off-topic by amWhy, Masacroso, Clarinetist, RRL, user10354138Dec 7 at 2:40

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• What sum will come next? – gammatester Dec 6 at 22:57
• with a software and a computer? Wolfram Mathematica gives the approximate result of $0.880734$ – Masacroso Dec 6 at 23:18
• $\displaystyle\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{(2n+1)^{2}}=8Li_4(\frac{1}{2})+\frac{\pi^2}{6}ln^22+\frac{1}{3}ln^42-\frac{61}{1440}{\pi^4}$ – user178256 Dec 7 at 8:49
• How can this be proven? – Isak Dec 7 at 10:23

$$\sum_{n\geq 1}H_n^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}$$ hence the explicit computation of your series is equivalent to the computation of $$\int_{0}^{1}\frac{\log^2(1-x^2)+\text{Li}_2(x^2)}{1-x^2}\log(x)\,dx$$ which can probably be tackled through Fourier(-Legendre) series. The simple part is $$\int_{0}^{1}\frac{\log^2(1-x^2)\log(x)}{1-x^2}\,dx = -\frac{\pi^4}{24}-\frac{\pi^2}{2}\log^2(2)+7\zeta(3)\log(2).$$
• Isn't this summing $H_n^2/(2n+1)$ as opposed to $H_n^2/(2n+1)^2$? – Badam Baplan Dec 6 at 23:33
• @BadamBaplan: it is summing the right thing, since $\int_{0}^{1}x^{2n}\log(x)\,dx = -\frac{1}{(2n+1)^2}$. – Jack D'Aurizio Dec 6 at 23:36
• @Isak: the difficult part is essentially $\sum_{n\geq 1}\frac{H_n^{(2)}}{(2n+1)^2}$ which appears to be manageable by summation by parts. – Jack D'Aurizio Dec 6 at 23:39