TITLE: Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)
QUESTION [7 upvotes]: This question arose from Amdeberhan's question, 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?

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.

REPLY [9 votes]: This is a special case of Watson's Theorem
$$\def\h{\frac{1}{2}}
\def\g#1{\Gamma(#1)\,}
{}_3F_2\left({a,\ b,\ c\atop\h a+\h b+\h, 2c }\biggm| 1 \right)
  =\frac{\g\h\g{c+\h}\g{\h a+\h b +\h}\g{c+\h -\h a -\h b}}
    {\g{\h a +\h} \g{\h b+\h} \g{c+\h -\h a} \g{c+\h -\h b}}.
$$