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.

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,  $\displaystyle \binom{2 N}{N}\sim \frac{4^N}{\sqrt{\pi N}}$.
