MHB Infinite Sums Involving cube of Central Binomial Coefficient

AI Thread Summary
The discussion centers on proving two infinite sums involving the cube of the central binomial coefficient. The first sum, with alternating signs, equals a specific expression involving the Gamma function, while the second sum, without alternating signs, yields a different expression also related to the Gamma function. Participants confirm that elliptic integrals or functions are necessary for solving these sums. References to specific equations on a linked page are suggested for further assistance. The conversation emphasizes the mathematical complexity and the role of special functions in deriving the results.
Shobhit
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$$
\begin{align*}
\sum_{n=0}^\infty \binom{2n}{n}^3 \frac{(-1)^n}{4^{3n}} &= \frac{\Gamma\left(\frac{1}{8}\right)^2\Gamma\left(\frac{3}{8}\right)^2}{2^{7/2}\pi^3} \tag{1}\\
\sum_{n=0}^\infty \binom{2n}{n}^3 \frac{1}{4^{3n}}&= \frac{\pi}{\Gamma \left(\frac{3}{4}\right)^4}\tag{2}
\end{align*}
$$

$\Gamma(z)$ denotes the Gamma Function.
 
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Am I right in assuming Elliptic integrals/functions are required for this one, Shobhit...?
 
DreamWeaver said:
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.
 
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