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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?

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  • $\begingroup$ Are you sure that first equality is correct? When I evaluate the LHS in Mathematica I get $\dfrac{\Gamma[1/4]}{\sqrt{2 \pi} \,\Gamma[3/4]}$ as the output. $\endgroup$ – A.E Jul 3 '13 at 14:57
  • $\begingroup$ @Orangutango: use the reflection formula $$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$ $\endgroup$ – Ron Gordon Jul 3 '13 at 15:04
  • $\begingroup$ @Orangutango: I think I did leave a factor of $\pi/2$ in there by mistake, however. $\endgroup$ – Ron Gordon Jul 3 '13 at 15:06
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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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  • $\begingroup$ I am using Mathematica notation, which does not square the argument in the square root. But what you do is fine by me. $\endgroup$ – Ron Gordon Jul 3 '13 at 15:14
  • $\begingroup$ Could you show how you expanded the integrand and got that sum? $\endgroup$ – Random Variable Jul 3 '13 at 15:50
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    $\begingroup$ Sure. You know the Taylor series of $(1-y^2)^{-1/2}$ is $$\sum_{n=0}^{\infty} \frac{1}{2^{2 n}} \binom{2 n}{n}$$ That then leaves you with that sum times $$\int_0^{\pi/2} d\theta \, \sin^{2 n}{\theta} = \frac{\pi}{2} \frac{1}{2^{2 n}} \binom{2 n}{n}$$ The last factor of $1/2^n$ comes from the factor of $1/2$ in the square root. $\endgroup$ – Ron Gordon Jul 3 '13 at 16:00
  • $\begingroup$ The question was how to arrive to the $\Gamma$'s involved result from the sum. In another words, how to evaluate the sum ?. $\endgroup$ – Felix Marin Jul 15 '14 at 1:35
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With reference to the first equation, I get the same constant with the reciprocal of the following integral:

$\frac{\sqrt{\pi}}{\Gamma\left(\frac{3}{4}\right)^2}=\frac{1}{{\int_{0}^\frac{\pi}{2}}\sqrt\sin(x)\sqrt\cos(x)\mathrm{d}x}$

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