Prove that $\sum_{n=0}^{\infty}\frac{\Gamma^2(n+1)}{\Gamma(2n+2)}=\frac{2\pi}{3^{3/2}}$ 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!
 A: 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
$$ \sum_{n\geq 0}\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\frac{d}{dz}\left.2\arcsin^2\tfrac{z}{2}\right|_{z=1}=\left.\frac{4\arcsin(z/2)}{\sqrt{4-z^2}}\right|_{z=1}=\frac{2\pi}{3\sqrt{3}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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.

A: 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
$$
\eqalign{
  & \sum\limits_{n = 0}^\infty  {{{\Gamma \left( {n + 1} \right)^{\,2} } \over {\Gamma \left( {2\left( {n + 1} \right)} \right)}}}
  = {}_2F_{\,1} \left( {\left. {\matrix{   {1,\;1}  \cr    {3/2}  \cr  } \;} \right|\;1/4} \right) =   \cr 
  &  = {{\arcsin \left( {\sqrt {1/4} } \right)} \over {\sqrt {1 - 1/4} \sqrt {1/4} }} = {{4\pi } \over {6\sqrt 3 }} = {{2\pi } \over {3\sqrt 3 }} \cr} 
$$
