Infinite Series $\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$ I am trying to find a closed form for this infinite series:
$$ S=\sum_{n=1}^{\infty}\frac{4^nH_n}{n^2{2n\choose n}}$$
Whith $H_n=\sum\limits_{k=1}^{n}\frac{1}{k}$ the harmonic numbers.
I found this integral representation of S:
$$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx$$
Sketch of a proof:
Recall the integral representation of the harmonic numbers: $H_n=\displaystyle\int_{0}^{1}\frac{1-x^n}{1-x}dx$
By plugging it into the definition of S and interchanging the order of summation between $\displaystyle\sum$ and $\displaystyle\int$ (justified by the uniform convergence of the function series $\displaystyle\sum\left(x\to\frac{4^n}{n^2{2n\choose n}}\frac{1-x^n}{1-x}\right)$, because $\forall x\in[0,1],\frac{1-x^n}{1-x}<n$), we get:
$$S=\int_{0}^{1}\frac{1}{1-x}\sum\limits_{n=1}^{\infty}\frac{4^n(1-x^n)}{n^2{2n\choose n}}dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
$$=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}\right)dx$$
Using the result $\displaystyle\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}=\frac{\pi^2}{2}$.
At that point, we will rely on the taylor series expansion of $\arcsin^2$:
$$\arcsin^2(x)=\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{4^n}{n^2{2n\choose n}}x^{2n}, |x|<1$$
Out of which we get $\displaystyle\sum\limits_{n=1}^{\infty}\frac{(4x)^n}{n^2{2n\choose n}}=2\arcsin^2\left(\sqrt{x}\right)$
So,
$$S=\int_{0}^{1}\frac{1}{1-x}\left(\frac{\pi^2}{2}-2\arcsin^2\left(\sqrt{x}\right)\right)dx$$
Which, through the substitution $u=\sqrt{x}$, gives the integral representation above.
But beyond that, nothing so far. I tried to use the integral representation of $\frac{H_n}{n}$ to switch the order of summation, but it didn't lead anywhere. Any suggestion?
 A: $$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx\overset{IBP}=-4\int_0^1 \frac{\arcsin x\ln(1-x^2)}{\sqrt{1-x^2}}dx$$
$$\overset{x=\sin t}=-8\int_0^\frac{\pi}{2} t \ln(\cos t)dt=8 \ln 2 \int_0^\frac{\pi}{2}t dt+8\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^\frac{\pi}{2} t\cos(2n t)dt$$
$$={\pi^2}\ln 2+2\sum_{n=1}^\infty \frac{1-(-1)^n}{n^3}=\boxed{\pi^2 \ln 2 +\frac72 \zeta(3)}$$
A: From here, we have 
$$\frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n=1}^\infty\frac{(2z)^{2n-1}}{n{2n \choose n}}$$
substitute $z=\sqrt{y}$, we get
$$\sum_{n=1}^\infty\frac{4^ny^n}{n{2n \choose n}}=2\sqrt{y}\frac{\arcsin\sqrt{y}}{\sqrt{1-y}}$$
Now multiply both sides by $-\frac{\ln(1-y)}{y}$ then integrate from $y=0$ to $1$ and using the fact that $-\int_0^1 y^{n-1}\ln(1-x)\ dy=\frac{H_n}{n}$, we get
\begin{align}
\sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n \choose 2}}&=-2\int_0^1\frac{\arcsin\sqrt{y}}{\sqrt{y}\sqrt{1-y}}\ln(1-y)\ dy\overset{\arcsin\sqrt{y}=x}{=}-8\int_0^{\pi/2}x\ln(\cos x)\ dx\\
&=-8\int_0^{\pi/2}x\left\{-\ln2-\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}\right\}\ dx\\
&=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x\cos(2nx) dx\\
&=\pi^2\ln2+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}\right)\\
&=\pi^2\ln2+2\pi\sum_{n=1}^\infty\frac{(-1)^n\sin(n\pi)}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^n\cos(n\pi)}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\\
&=\pi^2\ln2+0+2\sum_{n=1}^\infty\frac{(-1)^n(-1)^n}{n^3}-2\operatorname{Li}_3(-1)\\
&=\pi^2\ln2+2\zeta(3)-2\left(-\frac34\zeta(3)\right)\\
&=\pi^2\ln2+\frac72\zeta(3)
\end{align}
