Prove that $\sum_{n=0}^{\infty}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$

Question

I am seeking alternate proofs for $$\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$$ Here's mine:

Recall that, for $x\in(0,2)$, $$\frac1x=\sum_{n\geq0}(1-x)^n$$ Hence we have that, for $\frac{1-\sqrt5}2<x<\frac{1+\sqrt5}2$, $$\frac1{x^2-x+1}=\sum_{n\geq0}x^n(1-x)^n$$ Which gives $$ \begin{align} \int_0^1\frac{\mathrm dx}{x^2-x+1}=&\int_0^1\sum_{n\geq0}x^n(1-x)^n\mathrm dx\\ =&\sum_{n\geq0}\int_0^1x^n(1-x)^n\mathrm dx\\ =&\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)} \end{align} $$ That last step was from the definition of the Beta function: $$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Setting $I=\int_0^1\frac{\mathrm dx}{x^2-x+1}$, we complete the square: $$I=\int_0^1\frac{\mathrm dx}{(x-\frac12)^2+\frac34}$$ Preforming the substitution $x=\frac{1+3^{1/2}\tan u}2$, we have $$I=\frac{3^{1/2}}2\int_{-\pi/6}^{\pi/6}\frac{\sec^2u\ \mathrm du}{\frac34+\frac34\tan^2u}$$ And with a little simplification, we arrive at $$\sum_{n\geq0}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$$ Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!

Answer 1

A not-really-alternative approach:

$$ \sum_{n\geq 0}\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\sum_{n\geq 0}\frac{1}{(2n+1)\binom{2n}{n}}=\sum_{n\geq 1}\frac{2}{n\binom{2n}{n}} $$ and since $\sum_{n\geq 1}\frac{z^{2n}}{n^2\binom{2n}{n}}$ equals $2\arcsin^2\tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have $$ \sum_{n\geq 0}\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\frac{d}{dz}\left.2\arcsin^2\tfrac{z}{2}\right|_{z=1}=\left.\frac{4\arcsin(z/2)}{\sqrt{4-z^2}}\right|_{z=1}=\frac{2\pi}{3\sqrt{3}}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, $\ds{\Psi}$ is the Digamma Function: $\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ where $\ds{\Gamma}$ is the Gamma Function.

\begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over x^{2} - x + 1}} = \int_{0}^{1}{1 + x \over 1 + x^{3}}\,\dd x \\[5mm] = &\ \int_{0}^{1}{1 + x - x^{3} - x^{4}\over 1 - x^{6}}\,\dd x \\[5mm] = &\ {1 \over 6}\int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 6}\left\{\bracks{% \int_{0}^{1}{1 - x^{-1/3} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-2/3} \over 1 - x}\,\dd x} \right. \\[2mm] &\ + \left.\phantom{\left\{\,\right.} \bracks{\int_{0}^{1}{1 - x^{-1/6} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{-5/6} \over 1 - x}\,\dd x} \label{1}\tag{1} \right\} \\[5mm] = &\ {1 \over 6}\braces{\!% \bracks{\Psi\pars{2 \over 3} - \Psi\pars{1 \over 3}} + \bracks{\Psi\pars{5 \over 6} - \Psi\pars{1 \over 6}}\!}\label{2}\tag{2} \\[5mm] = &\ {1 \over 6}\bracks{\pi\cot\pars{\pi \over 3} + \pi\cot\pars{\pi \over 6}} = {1 \over 6}\pars{{\root{3} \over 3} + \root{3}}\pi \label{3}\tag{3} \\[5mm] = &\ \bbx{{2\root{3} \over 9}\,\pi} \approx 1.2092 \end{align}

\eqref{1} is evaluated with $\ds{\mathbf{\color{black}{6.3.22}}}$ A & S Identity. In \eqref{2} and \eqref{3}, I used the Euler Reflection Formula.

Answer 3

Still another way is through the Hypergeometric function.

Since the term of the sum is $$ t_{\,n} = {{\Gamma \left( {n + 1} \right)^{\,2} } \over {\Gamma \left( {2n + 2} \right)}} $$ and $$ \eqalign{ & t_{\,0} = {{\Gamma \left( 1 \right)^{\,2} } \over {\Gamma \left( 2 \right)}} = 1 \cr & {{t_{\,n + 1} } \over {t_{\,n} }} = {{\Gamma \left( {n + 2} \right)^{\,2} } \over {\Gamma \left( {2n + 4} \right)}}{{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 1} \right)^{\,2} }} = {{\left( {n + 1} \right)^{\,2} } \over {\left( {2n + 3} \right)\left( {2n + 2} \right)}} \cr & = {{\left( {n + 1} \right)\left( {n + 1} \right)} \over {\left( {n + 3/2} \right)}}{{1/4} \over {\left( {n + 1} \right)}} \cr} $$ then $$ \eqalign{ & \sum\limits_{n = 0}^\infty {{{\Gamma \left( {n + 1} \right)^{\,2} } \over {\Gamma \left( {2\left( {n + 1} \right)} \right)}}} = {}_2F_{\,1} \left( {\left. {\matrix{ {1,\;1} \cr {3/2} \cr } \;} \right|\;1/4} \right) = \cr & = {{\arcsin \left( {\sqrt {1/4} } \right)} \over {\sqrt {1 - 1/4} \sqrt {1/4} }} = {{4\pi } \over {6\sqrt 3 }} = {{2\pi } \over {3\sqrt 3 }} \cr} $$