This question arose from [Amdeberhan's question](https://mathoverflow.net/q/384145/11260), the evaluation of a double integral, which can be reduced to the evaluation of this series:
$$\sum \_{n=0}^{\infty } \frac{\Gamma \left(n+\frac{1}{2}\right)^2 \Gamma \left(n+\frac{s}{2}\right)}{\Gamma (n+1)^2 \Gamma (n+s)}=\frac{\pi ^2 2^{1-s} \Gamma \left(\frac{s}{2}\right)}{\left[\Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{s}{2} +\frac{1}{4}\right)\right]^2},\;\;{\rm Re}\,s>0.$$
The evaluation of the sum is Mathematica output.



> 
> Can someone enlighten me as to how this calculation proceeds?
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I went so far as to pay for Wolfram Alpha Pro, hoping that it would disclose the steps, but to no avail. What is even more frustrating is that for $s=1$ the right-hand-side is the square of a complete elliptic integral, which is also recognized immediately by Mathematica and was the original question in the cited post, so far without a conclusive answer.



Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)
Let's rewrite the given problem
$$\sum \_{n=0}^{\infty } \frac{\Gamma \left(n+\frac{1}{2}\right)^2 \Gamma \left(n+\frac{s}{2}\right)}{\Gamma (n+1)^2 \Gamma (n+s)}=\frac{\pi ^2 2^{1-s} \Gamma \left(\frac{s}{2}\right)}{\left[\Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{s}{2} +\frac{1}{4}\right)\right]^2}.\tag1$$
Change $s\rightarrow 2s$, then multiply equation (1) through by $\Gamma(s+1)$ to convert the problem to
$$\sum\_{n=0}^{\infty}16^{-n}\binom{2n}n^2\binom{2s+n-1}{s+n-1}^{-1}
=\frac{3\pi}{2^{2s+1}}\binom{s}{s-\frac34}\binom{s-1}{s-\frac34}.\tag2$$
Define the two functions
\begin{align\*}
F(s,n)&=16^{-n}\binom{2n}n^2\binom{2s+n-1}{s+n-1}^{-1}\frac{2^{2s+1}}{3\pi}\binom{s}{s-\frac34}^{-1}\binom{s-1}{s-\frac34}^{-1} \qquad \text{and} \\
G(s,n)&=\frac{n^2}{s(2s + n)}F(s,n).\end{align\*}
It is (mechanically) routine to verify that $F(s+1,n)-F(s,n)=G(s,n+1)-G(s,n)$. Now, sum both sides over all $n\in\mathbb{Z}\_{\geq0}$ resulting in the vanishing of the RHS. That means $f(s+1)=f(s)$ where $f(s)=\sum\_{n\geq0}F(s,n)$ or $f(s)$ is a constant. To determine the constant, check that $f(1)=1$ because
$$\sum\_{n=0}^{\infty}\frac1{16^n(n+1)}\binom{2n}n^2=\frac4{\pi}.$$
The latter evaluation can be justified in different ways.


**Remark.** This technique is called the Wilf-Zeilberger methodology. It could be what *Mathematica* or *Maple* is "hiding" from view. :-)


