<s>[INST] 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 ?
 [/INST] For $x \in [0,1]$ let
$$ f(x) = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n} x^{2n} \, . $$
Using the power series of $\arcsin$ we find
$$ x \frac{\mathrm{d}}{\mathrm{d} x} x \frac{\mathrm{d}}{\mathrm{d} x} f(x) = 4  \frac{\mathrm{d}}{\mathrm{d} x} [\arcsin(x) - x] = 4 \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] $$
for $x \in [0,1)$ . In particular,
$$ f'(1) = 4 \int \limits_0^1 \frac{1}{x} \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] \, \mathrm{d} x \stackrel{x=\sqrt{1-y^2}}{=} 4 \int \limits_0^1 \frac{\mathrm{d} y}{1+y} = 4 \ln(2) \, . $$
Now we can compute
\begin{align}
S &\equiv \sum \limits_{n=1}^\infty \frac{H_{2n} {2n \choose n}}{n^2 4^n} = \sum \limits_{n=1}^\infty \frac{{2n \choose n}}{n^2 4^n}  \int \limits_0^1 \frac{1-x^{2n}}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{f(1) - f(x)}{1-x} \, \mathrm{d} x \\
&= \int \limits_0^1 \frac{- \ln(1-x)}{x} x f'(x) \, \mathrm{d} x
= \operatorname{Li}_2 (1) f'(1) - 4 \int \limits_0^1 \frac{\operatorname{Li}_2 (x)}{x} \left[\frac{1}{\sqrt{1-x^2}} - 1 \right] \, \mathrm{d} x \\
&= \operatorname{Li}_2 (1) f'(1) + 4 \operatorname{Li}_3 (1) - 4 \int \limits_0^1 \frac{\operatorname{Li}_2 (x)}{x \sqrt{1-x^2}} \, \mathrm{d} x \equiv 4 \left[\frac{\pi^2}{6} \ln(2) + \zeta(3) - I\right] \, .
\end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
\begin{align}
I &= \int \limits_0^\infty t \int \limits_0^1 \frac{\mathrm{d} x}{(\mathrm{e}^t - x) \sqrt{1-x^2}} \, \mathrm{d} t \stackrel{(*)}{=} \int \limits_0^\infty \frac{t \left[\frac{\pi}{2} + \arcsin(\mathrm{e}^{-t})\right]}{\sqrt{\mathrm{e}^{2t}-1}} \, \mathrm{d} t \\
&\stackrel{\mathrm{e}^{-t} = \sin(u)}{=} \frac{1}{2} \int \limits_0^{\pi/2} -\ln[\sin(u)] (\pi + 2 u) \, \mathrm{d} u = \frac{1}{2} \int \limits_0^{\pi/2} u (\pi + u) \cot(u) \, \mathrm{d} u \\
&= \frac{1}{2} [\pi K_1^{(1)} + K_2^{(1)}] = \frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3) \, .
\end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ \boxed{S = \sum \limits_{n=1}^\infty \frac{H_{2n} {2n \choose n}}{n^2 4^n} = \frac{23}{4} \zeta(3) - \frac{5}{6} \pi^2 \ln(2)} \, . $$

Proof of $(*)$:
For $a \in [0,1]$ let
$$ g(a) = \int \limits_0^1 \frac{-\ln(1-a x)}{x \sqrt{1-x^2}} \, \mathrm{d} x= \sum \limits_{n=1}^\infty \frac{a^n}{n} \int \limits_0^{\pi/2} \sin^{n-1} (t) \, \mathrm{d} t \, .$$
Using Wallis' integrals we find
$$ g(a) = \frac{\pi}{2} \sum \limits_{k=0}^\infty \frac{{2k \choose k} a^{2k+1}}{4^k(2k+1)} + \frac{1}{4} \sum \limits_{m=1}^\infty \frac{4^k a^{2k}}{k^2 {2k \choose k}} = \frac{\pi}{2} \arcsin(a) + \frac{1}{2} \arcsin^2 (a) \, . $$
Therefore
$$ \int \limits_0^1 \frac{\mathrm{d} x}{(1-a x)\sqrt{1-x^2}} = g'(a) = \frac{\frac{\pi}{2} + \arcsin{a}}{\sqrt{1-a^2}} $$
holds for $a \in [0,1)$ .
</s>