# 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?

• 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. – A.E Jul 3 '13 at 14:57
• @Orangutango: use the reflection formula $$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$ – Ron Gordon Jul 3 '13 at 15:04
• @Orangutango: I think I did leave a factor of $\pi/2$ in there by mistake, however. – Ron Gordon Jul 3 '13 at 15:06

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.

$$\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\frac{1}{2} \sin^{2} x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2-\sin^{2}x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+ \cos^{2} x}} \ dx$$

$$= \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1+u^{2}}} \frac{du}{\sqrt{1-u^{2}}} = \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-u^{4}}} du$$

$$= \frac{\sqrt{2}}{4} \int_{0}^{1} t^{-\frac{3}{4}} (1-t)^{\frac{-1}{2}} \ dt = \frac{\sqrt{2}}{4} B \left( \frac{1}{4}, \frac{1}{2} \right) = \frac{\sqrt{2}}{4} \frac{\sqrt{\pi} \ \Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}$$

• I am using Mathematica notation, which does not square the argument in the square root. But what you do is fine by me. – Ron Gordon Jul 3 '13 at 15:14
• Could you show how you expanded the integrand and got that sum? – Random Variable Jul 3 '13 at 15:50
• 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. – Ron Gordon Jul 3 '13 at 16:00
• The question was how to arrive to the $\Gamma$'s involved result from the sum. In another words, how to evaluate the sum ?. – Felix Marin Jul 15 '14 at 1:35

With reference to the first equation, I get the same constant with the reciprocal of the following integral:

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


\begin{align} &\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{1 \over 2^{5n}} {2n \choose n}^{2}} = \sum_{n = 0}^{\infty}{1 \over 2^{5n}} \bracks{{-1/2 \choose n}\pars{-4}^{n}}^{2} \\[5mm] = &\ \sum_{n = 0}^{\infty}{1 \over 2^{n}}{-1/2 \choose n} \oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{-1/2} \over z^{n +1}} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1}{1 \over z\root{z + 1}} \sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{1 \over 2z}^{n} \,{\dd z \over 2\pi\ic} \\[5mm] = &\ \oint_{\verts{z}\ =\ 1}{1 \over z\root{z + 1}} \pars{1 + {1 \over 2z}}^{-1/2}\,{\dd z \over 2\pi\ic} = \oint_{\verts{z}\ =\ 1}{\root{z} \over z\root{z + 1}\root{z + 1/2}} \,{\dd z \over 2\pi\ic} \\[5mm] = & -\int_{-1}^{-1/2}{\root{-z}\ic \over z\root{z + 1}\root{-z - 1/2}\ic}\,{\dd z \over 2\pi\ic} -\int_{-1/2}^{0}{\root{-z}\ic \over z\root{z + 1}\root{z + 1/2}}\,{\dd z \over 2\pi\ic} \\[2mm] & - \int_{0}^{-1/2}{\root{-z}\pars{-\ic} \over z\root{z + 1}\root{z + 1/2}}\,{\dd z \over 2\pi\ic} -\int_{-1/2}^{-1}{\root{-z}\pars{-\ic} \over z\root{z + 1}\root{-z - 1/2}\pars{-\ic}}\,{\dd z \over 2\pi\ic} \\[5mm] = & {1 \over 2\pi\ic}\int_{1/2}^{1}{\root{z} \over z\root{1 - z}\root{z - 1/2}}\,\dd z + {1 \over 2\pi}\int_{0}^{1/2}{\root{z} \over z\root{1 - z}\root{1/2 - z}}\,\dd z \\[2mm] & + {1 \over 2\pi}\int_{0}^{1/2}{\root{z} \over z\root{1 - z}\root{1/2 - z}}\,\dd z - {1 \over 2\pi\ic}\int_{1/2}^{1} {\root{z} \over z\root{1 - z}\root{z - 1/2}}\,\dd z \\[5mm] = &\ {1 \over \pi}\ \underbrace{\int_{0}^{1/2}{\root{z} \over z\root{1 - z}\root{1/2 - z}}\,\dd z} _{\ds{2\root{2\pi}\Gamma\pars{5/4} \over \Gamma\pars{3/4}}}\ =\ {2\root{2\pi} \over \pi}\,{\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{3/4}} \\[5mm] = &\ {\root{2\pi} \over 2\pi}\,{\Gamma\pars{3/4}\Gamma\pars{1/4} \over \Gamma^{2}\pars{3/4}} = {\root{2\pi} \over 2\pi}\,{\pi/\sin\pars{\pi/4} \over \Gamma^{2}\pars{3/4}} = {\root{2\pi} \over 2\pars{1/\root{2}}}\,{1 \over \Gamma^{2}\pars{3/4}} \\[5mm] = &\ \bbox[10px,#ffd,border:2px groove navy] {\Large{\root{\pi} \over \Gamma^{2}\pars{3/4}}} \approx 1.1803 \end{align}

The last integral is evaluated with the replacement $$\ds{z = \sin^{2}\pars{\theta}}$$ which follows the $${\tt @Random Variable}$$ answer procedure. The identity $$\ds{{2n \choose n} = {-1/2 \choose n}\pars{-4}^{n}}$$ is derived in this link.