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Infinite Sums Involving cube of Central Binomial Coefficient

Shobhit

Member
Nov 12, 2013
23
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
Sep 16, 2013
337
Am I right in assuming Elliptic integrals/functions are required for this one, Shobhit...?
 

Shobhit

Member
Nov 12, 2013
23
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.