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?

  • $\begingroup$ 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

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
  • 2
    $\begingroup$ 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

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


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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{n = 0}^{\infty}{1 \over 2^{5n}}{2n \choose n}^{2} ={\root{\pi} \over \Gamma^{\,2}\pars{3/4}}}$

\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.

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