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 ?
 A: 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 $a \in [0,1)$ .
A: Using the fact that $$\int_0^1x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$$
multiply both sides by $\ \displaystyle-\frac{2}{n4^n}{2n \choose n}$ then take the sum, we get,
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n}}{n^24^n}{2n \choose n}=-2\int_0^1\frac{\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(x^2)^n}{n4^n}{2n \choose n}\right)\ dx
\end{align}
I managed here to prove: $$\quad\displaystyle\sum_{n=1}^\infty \frac{x^n}{n4^n}{2n \choose n}=-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2$$
which follows:
\begin{align}
S&=4\underbrace{\int_0^1\frac{\ln(1-x)\tanh^{-1}{\sqrt{1-x^2}}}{x}\ dx}_{\text{IBP}}+4\int_0^1\frac{\ln(1-x)\ln x}{x}\ dx-4\ln2\int_0^1\frac{\ln(1-x)}{x}\ dx\\
&=-4\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx+4(\zeta(3))-4\ln2(-\zeta(2))\\
&=-4\left(\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)\right)+4\zeta(3)+\frac{2}{3}\pi^2\ln(2)\\
&\boxed{=\frac{23}4\zeta(3)-\frac{5}{6}\pi^2\ln2}
\end{align}
Credit goes to ComplexYetTrivial for nicely evaluating $\ \displaystyle\int_0^1\frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}\ dx=\frac{3}{8}\pi^2 \ln(2) - \frac{7}{16} \zeta(3)$ 
A: 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:
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=\frac{23}{4}\zeta \left(3\right)-5\ln \left(2\right)\zeta \left(2\right)$$
