Infinite Sums Involving cube of Central Binomial Coefficient

[Shobhit]

Show that

$$
\begin{align*}
\sum_{n=0}^\infty \binom{2n}{n}^3 \frac{(-1)^n}{4^{3n}} &= \frac{\Gamma\left(\frac{1}{8}\right)^2\Gamma\left(\frac{3}{8}\right)^2}{2^{7/2}\pi^3} \tag{1}\\
\sum_{n=0}^\infty \binom{2n}{n}^3 \frac{1}{4^{3n}}&= \frac{\pi}{\Gamma \left(\frac{3}{4}\right)^4}\tag{2}
\end{align*}
$$

$\Gamma(z)$ denotes the Gamma Function.

 

[DreamWeaver]

Am I right in assuming Elliptic integrals/functions are required for this one, Shobhit...?

 

[Shobhit]

Am I right in assuming Elliptic integrals/functions are required for this one, Shobhit...?

Yes, that is how they can be solved. You may have to use equations (3) and (6) on this page.