Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k}\text{ } \text{ } \text{} \text{ } \text{ }\text{the } n^{th} \text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?

Here is my way of evaluating this sum also offering a different way to calculate that polylogarithmic integral.

$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}$$


First let's consider the following central binomial coefficient generating function.
$$\sum _{k=1}^{\infty }\frac{x^{2k}}{k\:4^k}\binom{2k}{k}=-2\ln \left(1+\sqrt{1-x^2}\right)+2\ln \left(2\right)$$
$$-2\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}\int _0^1x^{2k-1}\ln \left(1-x\right)\:dx=4\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx$$
$$-4\ln \left(2\right)\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=4\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx+4\ln \left(2\right)\zeta \left(2\right)$$

$$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx=\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx-\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x\sqrt{1-x^2}}\:dx$$
$$=\zeta \left(3\right)-\int _0^{\frac{\pi }{2}}\csc \left(x\right)\operatorname{Li}_2\left(\sin \left(x\right)\right)\:dx=\zeta \left(3\right)-\int _0^1\frac{\operatorname{Li}_2\left(\frac{2t}{1+t^2}\right)}{t}dt$$
$$=\zeta \left(3\right)+2\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(t\right)\ln \left(1-t\right)}{t}\:dt$$
$$+\int _0^1\frac{\ln \left(t\right)\ln \left(1+t^2\right)}{t}\:dt+2\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt$$
$$=\zeta \left(3\right)+4\underbrace{\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt}_{I}+\frac{1}{4}\underbrace{\int _0^1\frac{\ln ^2\left(1+t\right)}{t}\:dt}_{\frac{1}{4}\zeta \left(3\right)}$$
$$-2\underbrace{\int _0^1\frac{\ln \left(t\right)\ln \left(1-t\right)}{t}\:dt}_{\zeta \left(3\right)}+\underbrace{\int _0^1\frac{\ln \left(t\right)\ln \left(1+t^2\right)}{t}\:dt}_{-\frac{3}{16}\zeta \left(3\right)}$$
The integral $I$ can be found evaluated elegantly in the book (Almost) Impossible Integrals, Sums, and Series through pages $\#98,\#99,\#100$, making use of it's result we have:
$$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx=\frac{23}{16}\zeta \left(3\right)-\frac{9}{4}\ln \left(2\right)\zeta \left(2\right)$$
And in the process we also proved that:
$$\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x\sqrt{1-x^2}}\:dx=-\frac{7}{16}\zeta \left(3\right)+\frac{9}{4}\ln \left(2\right)\zeta \left(2\right)$$

Collecting the results we finally have:
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=\frac{23}{4}\zeta \left(3\right)-5\ln \left(2\right)\zeta \left(2\right)$$
