The sum of an hydrogen atom related infinite series Let $$a_n=\frac{1}{n+\frac{1}{2}}\left [\frac{\Gamma(n)}{\Gamma(n+\frac{1}{2})}\right]^2,$$ and $$b_n=\frac{1}{n^2}.$$
On the ground of hydrogen atom quantum physics, it was shown in http://arxiv.org/abs/1510.07813 (Quantum Mechanical Derivation of the Wallis Formula for $\pi$, by T. Friedmann and C. R. Hagen) that
$$\lim\limits_{n\to\infty}\frac{a_n}{b_n}=1.$$
Therefore, since $$\sum\limits_{n=1}^\infty b_n=\frac{\pi^2}{6},$$
the series $\sum\limits_{n=1}^\infty a_n$ is convergent, according to the limit comparison test. Alternatively we can use direct comparison test, because $a_n\le b_n$ for all $n$ (also from physics).
Can the sum $\sum\limits_{n=1}^\infty a_n$ be calculated explicitly?
 A: A more general identity is furnished by
$$\sum_{n=1}^N\frac{\Gamma(n)\Gamma(n+k)}{\Gamma(n+\frac12)\Gamma(n+k+\frac32)}
=\frac4{2k+1}\left[\frac{\Gamma(N+1)\Gamma(N+k+1)}{\Gamma(N+\frac12)\Gamma(N+k+\frac32)}-\frac{\Gamma(k+1)}{\Gamma(k+\frac32)\sqrt{\pi}}\right],$$
which is provable by induction on $N$. Now, taking the limit $N\rightarrow\infty$ leads to
$$\sum_{n=1}^{\infty}\frac{\Gamma(n)\Gamma(n+k)}{\Gamma(n+\frac12)\Gamma(n+k+\frac32)}=\frac4{2k+1}\left[1-\frac{\Gamma(k+1)}{\Gamma(k+\frac32)\sqrt{\pi}}\right].$$
To get the solution to the OP's original problem, take $k=0$.
A: Incidentally, note that the series telescopes, as
$$a_n={\Gamma(n)^2\over\Gamma(n+1/2)\Gamma(n+3/2)}=-4\big[(n-1/2)(n+1/2)-n^2\big]{\Gamma(n)^2\over\Gamma(n+1/2)\Gamma(n+3/2)}=$$
$$=-{4\,\Gamma(n)^2\over\Gamma(n-1/2)\Gamma(n+1/2)}+{4\,\Gamma(n+1)^2\over\Gamma(n+1/2)\Gamma(n+3/2)},$$
whence the partial and the limit sum given in previous answers.
A: Maple also gets $4 - 8/\pi$.   This comes from an explicit formula for the partial sums:
$$ \sum_{n=1}^N a_n = {\frac { \left( 4\,N+2 \right)  \left( \Gamma \left( N+1 \right) 
 \right) ^{2}}{ \left( \Gamma \left( N+3/2 \right)  \right) ^{2}}}-\frac{8}
{\pi}
$$
which is readily verified by mathematical induction.
A: Mathematica says $$\frac{4 (\pi -2)}{\pi },$$ which is borne out by numerical approximation.
