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!

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