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I've been trying to compute the following series for quite a while :

$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$$ where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$ are the generalized harmonic numbers of order 2.

I've already successfully computed the value of a lot of similar-looking series involving harmonic numbers and central binomial coefficients (by using Abel's transformation, finding the generating function...), and they all had closed forms in terms of the Riemann Zeta function, so I'm confident this one too has a nice closed form ; but it's a tough guy that resists my previous methods.

By numerical test, I suspect that the value is $\frac{3}{2}\zeta(3)$.

Any idea for derivation ?

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    $\begingroup$ What a fitting name! $\endgroup$ – Frank W. Aug 26 '18 at 23:10
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Since

$$ \zeta(2)-H_n^{(2)} = \int_{0}^{1} x^n \frac{-\log x}{1-x}\,dx $$ $$ \sum_{n\geq 1}\frac{x^n}{n4^n}\binom{2n}{n} = 2\log(2)-2\log(1+\sqrt{1-x}) $$ the computation of your series boils down to the computation of the integral

$$ \int_{0}^{1}\frac{\log(x)\log(1+\sqrt{1-x})}{1-x}\,dx=\int_{0}^{1}\frac{\log(1-x)\log(1+\sqrt{x})}{x}\,dx $$ or the computation of the integral $$ \int_{0}^{1}\frac{\left[\log(1-x)+\log(1+x)\right]\log(1+x)}{x}\,dx $$ where $$ \int_{0}^{1}\frac{\log^2(1+x)}{x}=\frac{\zeta(3)}{4},\qquad \int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}\,dx =-\frac{5\,\zeta(3)}{8}$$ are simple to prove by Maclaurin series and Euler sums. Your conjecture is correct.

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