# Infinite Sums Involving cube of Central Binomial Coefficient

#### Shobhit

##### Member
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

##### Well-known member
Am I right in assuming Elliptic integrals/functions are required for this one, Shobhit...?

#### Shobhit

##### Member
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.