<s>[INST] Sum of squares of binomial coefficients I came across the following sum in reference to this question
$$\sum_{n=0}^{\infty} \frac{1}{2^{5 n}} \binom{2 n}{n}^2 = \frac{\sqrt{\pi}}{\Gamma \left( \frac{3}{4}\right)^2}$$
The sum on the left was generated from expanding the square root in the integrand of the following elliptic integral:
$$K\left( \frac{1}{2}\right) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac12 \sin^2{\theta}}} $$
For the life of me, I cannot figure out how to evaluate this sum directly.  Mathematica has no problem in doing so.  Can someone point the way?
 [/INST] I'm assuming you meant $K \left(\frac{1}{\sqrt{2}} \right)$ because that's what you have on the right side of the equation.
$$ \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\frac{1}{2} \sin^{2} x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2-\sin^{2}x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+ \cos^{2} x}} \ dx$$
$$= \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1+u^{2}}} \frac{du}{\sqrt{1-u^{2}}}  = \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-u^{4}}} du $$ 
$$= \frac{\sqrt{2}}{4} \int_{0}^{1} t^{-\frac{3}{4}} (1-t)^{\frac{-1}{2}} \ dt = \frac{\sqrt{2}}{4} B \left( \frac{1}{4}, \frac{1}{2} \right) = \frac{\sqrt{2}}{4} \frac{\sqrt{\pi} \ \Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}$$
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