<s>[INST] A gamma summation: $\sum_{n=0}^{\infty} \frac{2}{\Gamma ( a + n) \Gamma ( a - n )} = \frac{2^{2a-2}}{\Gamma ( 2a - 1 )} + \frac{1}{\Gamma^2 (a)}$ Let $a \notin \mathbb{Z}$ and $a \neq \frac{1}{2}$. Prove that
$$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$$
Attempt
Using the fact that
\begin{align*} 
\frac{1}{\Gamma\left ( a+x \right ) \Gamma \left ( \beta - x \right )} &= \frac{1}{\left ( a+x-1 \right )! \left ( \beta-x-1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \frac{\left ( a + \beta-2 \right )!}{\left ( a + x -1 \right )! \left ( \beta - x -1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \binom{a + \beta - 2}{a + x -1} 
\end{align*}
the question really boils down to the sum
$$\mathcal{S} = \sum_{n=0}^{\infty} \binom{2a-2}{a+n-1}$$
To this end,
\begin{align*}
 \sum_{n=0}^{\infty} \binom{2a-1}{a+n-1} &=\frac{1}{2\pi i} \sum_{n=0}^{\infty} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a+n}}\, \mathrm{d}z \\ 
 &= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1 + z \right )^{2a-1}}{z^a} \sum_{n=0}^{\infty} \frac{1}{z^n} \, \mathrm{d}z  \\ 
 &= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a-1} \left ( z-1 \right )} \, \mathrm{d}z
\end{align*}
using the handy identity $\displaystyle \binom{n}{k} = \frac{1}{2\pi i } \oint \limits_{\gamma} \frac{\left ( 1+z \right )^n}{z^{k+1}} \, \mathrm{d}z$. I think I'm on the right track, but I'm having a difficult time evaluating the last contour integral. Any help?
 [/INST] If you enjoy special functions
$$f(x)=\sum_{n=0}^{\infty} \frac{2\,x^n}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2 }{\Gamma (a)^2}\,\, _2F_1(1,1-a;a;-x)$$
$$\, _2F_1(1,1-a;a;-1)=\frac{1}{2} \left(1+\sqrt{\pi }\frac{ \Gamma (a)}{\Gamma
   \left(a-\frac{1}{2}\right)}\right)$$
$$f(1)=\frac{\sqrt{\pi }}{\Gamma \left(a-\frac{1}{2}\right) \Gamma (a)}+\frac{1}{\Gamma (a)^2}$$
Now, work the first term to get the desired result.
</s>