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