<s>[INST] 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?
 [/INST] $$S=2\int_{0}^{1}\frac{x}{1-x^2}\left(\frac{\pi^2}{2}-2\arcsin^2(x)\right)dx\overset{IBP}=-4\int_0^1 \frac{\arcsin x\ln(1-x^2)}{\sqrt{1-x^2}}dx$$
$$\overset{x=\sin t}=-8\int_0^\frac{\pi}{2} t \ln(\cos t)dt=8 \ln 2 \int_0^\frac{\pi}{2}t dt+8\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^\frac{\pi}{2} t\cos(2n t)dt$$
$$={\pi^2}\ln 2+2\sum_{n=1}^\infty \frac{1-(-1)^n}{n^3}=\boxed{\pi^2 \ln 2 +\frac72 \zeta(3)}$$
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