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?
 A: If $a\notin\mathbb{Z}$ then the sum converges if and only if $\color{red}{a>1/2}$, which is obtained from $$\lim_{x\to+\infty}\frac{x^\lambda\Gamma(x)}{\Gamma(\lambda+x)}=1\implies\lim_{n\to\infty}\frac{(-1)^n\color{red}{n^{2a-1}}}{\Gamma(a+n)\Gamma(a-n)}=\frac{\sin a\pi}{\pi}$$ using the reflection formula for $\Gamma(a-n)$.
In this case, a possible solution is to apply the Poisson summation formula $$\sum_{n\in\mathbb{Z}}f(n)=\sum_{n\in\mathbb{Z}}\hat{f}(n),\qquad\hat{f}(y):=\int_{-\infty}^\infty f(x)e^{-2i\pi xy}\,dx$$ to $f(x)=\cos^{2(a-1)}\pi x$ for $|x|<1/2$ (and $f(x)=0$ elsewhere); then we get $$\hat{f}(y)=\frac{2^{2-2a}\Gamma(2a-1)}{\Gamma(a+y)\Gamma(a-y)}$$ as a "reciprocal beta" integral (see DLMF or e.g. this question), and the result follows.
A: To supplement Claude Leibovici's answer, I will use Euler's integral representation of the Gaussian hypergeometric  function to show that $$_2F_1(1,1-a;a;-1)=\frac{1}{2} \left(\frac{2^{2a-2} \, \Gamma^{2}(a)}{\Gamma(2a-1)} +1 \right) $$ for at least $a >1$.  This is where the series $\sum_{n=0}^{\infty} \frac{2}{\Gamma ( a + n) \Gamma ( a - n )} $ converges absolutely.
$ \begin{align} _2F_1(1,1-a;a;-1) &= \, _2F_1(1-a,1;a;-1) \\ &= \frac{1}{B(1,a-1)} \int_{0}^{1} (1-x)^{a-2} (1+x)^{a-1} \, \mathrm dx \\ &= (a-1) \int_{0}^{1} (1-x)^{a-2}(1+x)^{a-2}(1+x) \, \mathrm dx\\ &= (a-1) \left( \int_{0}^{1} (1-x^{2})^{a-2} \, \mathrm dx +\int_{0}^{1} x (1-x^{2})^{a-2} \, \mathrm dx \right) \\ &= (a-1) \left( \frac{1}{2} \int_{0}^{1} (1-u)^{a-2} u^{-1/2} \, \mathrm du + \frac{1}{2}\int_{0}^{1} v^{a-2} \, \mathrm dv \right) \\ &= (a-1) \left( \frac{1}{2} \, B \left(\frac{1}{2},a-1\right) + \frac{1}{2(a-1)} \right) \\ &= \frac{a-1}{2} \left(\frac{ \sqrt{\pi} \,   \Gamma(a-1)}{\Gamma \left(a- \frac{1}{2}\right)}+\frac{1}{a-1} \right) \\ &\overset{(1)}= \frac{a-1}{2} \left(\frac{\sqrt{\pi} \, \Gamma(a) 2^{2(a-1/2)-1} \Gamma \left(a-\frac{1}{2}+\frac{1}{2} \right)}{(a-1) \Gamma \left( 2\left(a-\frac{1}{2}\right)\right) \sqrt{\pi}}+\frac{1}{a-1} \right) \\ &= \frac{1}{2} \left(\frac{2^{2a-2} \Gamma^{2}(a)}{\Gamma(2a-1)} +1 \right). \end{align}$

$(1)$ Legendre duplication formula
A: 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.
