A sum of series with the inverse squared central binomial coefficient 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.
 A: An excellent answer was already given (the chosen one), but good to have more ways in place.


A solution by Cornel Ioan Valean
Instead of calculating all three series separately, we might try to calculate them all at once. So, we have that
$$2\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^3 \binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{2^{4 n} H_n^{(2)}}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty }\frac{2^{4n} (4n^2-1+n^2 H_n^{(2)})}{\displaystyle n^4 (2 n+1) \binom{2 n}{n}^2}=\sum _{n=1}^{\infty }\frac{2^{4n} (4-1/n^2+ H_n^{(2)})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty }\frac{2^{4n}(4-1/n^2+ H_n^{(2)}\color{blue}{+(4 n^2-1) H_{n-1}^{(2)}}-\color{blue}{(4 n^2-1) H_{n-1}^{(2)}})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty }\frac{2^{4n}(\color{red}{4n^2H_n^{(2)}}-\color{blue}{(4 n^2-1) H_{n-1}^{(2)}})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty}\left(\frac{2^{4n+2}H_n^{(2)}}{\displaystyle (2n+1) \binom{2 n}{n}^2}-\frac{2^{4n}(2n-1)H_{n-1}^{(2)} }{\displaystyle n^2\binom{2 n}{n}^2}\right)$$
$$=\lim_{N\to\infty}\sum _{n=1}^{N}\left(\frac{2^{4n+3}H_n^{(2)}}{\displaystyle (n+1) \binom{2 n+2}{n+1}\binom{2 n}{n}}-\frac{2^{4n-1}H_{n-1}^{(2)} }{\displaystyle n\binom{2 n}{n}\binom{2 n-2}{n-1}}\right)$$
$$=\lim_{N\to\infty}\frac{2^{4N+3}H_N^{(2)}}{\displaystyle (N+1) \binom{2 N+2}{N+1}\binom{2 N}{N}}=\frac{\pi^3}{3},$$
where we used the asymptotic form of the central binomial coefficient,  $\displaystyle \binom{2 N}{N}\sim \frac{4^N}{\sqrt{\pi N}}$.
