TITLE: A sum of series with the inverse squared central binomial coefficient
QUESTION [9 upvotes]: A nice challenge by Cornel Valean:
Show that
$$2\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^3 \binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{2^{4 n} H_n^{(2)}}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}=\frac{\pi^3}{3}.$$
I have to say that I am not experienced in series involving squared central binomial coefficient, so I leave it for people who are experts in such series.
All approaches are appreciated. Thank you.

REPLY [4 votes]: Here's a simple way to prove (with the manipulation of well-known generating functions and integrals) that:

$$S=2\sum _{k=1}^{\infty }\frac{16^k}{k^3\binom{2k}{k}^2}-\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}+\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}=\frac{\pi ^3}{3}.$$

First let's make use of the following generating function:
$$\sum _{k=1}^{\infty }\frac{4^kx^{2k}}{k^2\binom{2k}{k}}=2\arcsin ^2\left(x\right)$$
$$\sum _{k=1}^{\infty }\frac{4^ky^{2k}}{k^3\binom{2k}{k}}=4\int _0^y\frac{\arcsin ^2\left(x\right)}{x}\:dx$$
$$2\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}\int _0^{\frac{\pi }{2}}\sin ^{2k-1}\left(y\right)\:dy=8\int _0^{\frac{\pi }{2}}\csc \left(y\right)\int _0^{\sin \left(y\right)}\frac{\arcsin ^2\left(x\right)}{x}\:dx\:dy$$
$$\boxed{\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}=8\int _0^{\frac{\pi }{2}}\csc \left(y\right)\int _0^yx^2\cot \left(x\right)\:dx\:dy}.$$
Now let's jump into the $3$rd series by first considering the same generating function used previously:
$$\sum _{k=1}^{\infty }\frac{4^kx^{2k}}{k^2\binom{2k}{k}}=2\arcsin ^2\left(x\right)$$
$$\sum _{k=1}^{\infty }\frac{4^k}{k^3\left(2k+1\right)\binom{2k}{k}}t^{2k}=4\int _0^t\left(2\frac{\sqrt{1-y^2}\arcsin \left(y\right)}{y^2}-2\frac{1}{y}+\frac{\arcsin ^2\left(y\right)}{y}\right)\:dy$$
$$\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}=8\int _0^{\frac{\pi }{2}}\csc \left(t\right)\int _0^t\left(2x\cot ^2\left(x\right)-2\cot \left(x\right)+x^2\cot \left(x\right)\right)\:dx\:dt$$
$$\boxed{\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}=16\int _0^{\frac{\pi }{2}}\csc \left(t\right)\left(1-t\cot \left(t\right)\right)\:dt-8\int _0^{\frac{\pi }{2}}t^2\csc \left(t\right)\:dt+\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}}.$$
Lastly consider:
$$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k\binom{2k}{k}}x^{2k}-\sum _{k=1}^{\infty }\frac{4^k}{k^3\binom{2k}{k}}x^{2k}=\frac{4}{3}\frac{x\arcsin ^3\left(x\right)}{\sqrt{1-x^2}}$$
$$\sum _{k=1}^{\infty }\frac{4^kH_k^{\left(2\right)}}{k\left(2k+1\right)\binom{2k}{k}}y^{2k+1}-\sum _{k=1}^{\infty }\frac{4^k}{k^3\left(2k+1\right)\binom{2k}{k}}y^{2k+1}=\frac{4}{3}\int _0^y\frac{x\arcsin ^3\left(x\right)}{\sqrt{1-x^2}}\:dx$$
$$\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}-\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}=\frac{8}{3}\int _0^{\frac{\pi }{2}}\csc ^2\left(y\right)\int _0^{\sin \left(y\right)}\frac{x\arcsin ^3\left(x\right)}{\sqrt{1-x^2}}\:dx\:dy$$
$$\boxed{\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}=\frac{8}{3}\int _0^{\frac{\pi }{2}}x^3\cos \left(x\right)\:dx+\sum _{k=1}^{\infty }\frac{16^k}{k^4\left(2k+1\right)\binom{2k}{k}^2}}.$$
And finally from here we know that:
$$\sum _{k=1}^{\infty }\frac{16^k}{k^3\binom{2k}{k}^2}=4\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\:dx,$$
plugging these results into the main expression we obtain:
$$S=\require{cancel}\cancel{8\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\:dx}-\cancel{\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}}+\frac{8}{3}\int _0^{\frac{\pi }{2}}x^3\cos \left(x\right)\:dx$$
$$\require{cancel}+16\int _0^{\frac{\pi }{2}}\csc \left(t\right)\left(1-t\cot \left(t\right)\right)\:dt-\cancel{8\int _0^{\frac{\pi }{2}}t^2\csc \left(t\right)\:dt}+\cancel{\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}}.$$
Therefore:
$$S=2\sum _{k=1}^{\infty }\frac{16^k}{k^3\binom{2k}{k}^2}-\sum _{k=1}^{\infty }\frac{16^k}{k^4\binom{2k}{k}^2}+\sum _{k=1}^{\infty }\frac{16^kH_k^{\left(2\right)}}{k^2\left(2k+1\right)\binom{2k}{k}^2}$$
$$=\frac{8}{3}\int _0^{\frac{\pi }{2}}x^3\cos \left(x\right)\:dx+16\int _0^{\frac{\pi }{2}}\csc \left(t\right)\left(1-t\cot \left(t\right)\right)\:dt=\frac{\pi ^3}{3},$$
and that is it.