<s>[INST] Limit of series involving ratio of two factorials $$
  \sum^{\infty}_{j=0} \frac{(j!)^2}{(2j)!} = \frac{2 \pi \sqrt{3}}{27}+\frac{4}{3}
$$
The above series is in a homework sheet. We're not expected to find the limit, just prove its convergence. That's easy, but since we were given the limit, it got me thinking about how to find such a limit.
If anyone could point me in the right direction, I'd be happy to discover it on my own, but after a few hours of searching, I don't feel much closer. 
 [/INST] Notice that:
$$
    \frac{j!^2}{(2j)!} = (2j+1) \operatorname{Beta}(j+1,j+1) = (2j+1) \int_0^1 t^j (1-t)^j \mathrm{d}t
$$
Now, interchanging the summation and the integration:
$$
  \sum_{j=0}^\infty \frac{j!^2}{(2j)!}  = \int_0^1 \sum_{j=0}^\infty (2j+1)(t(1-t))^j \mathrm{d} t = \int_0^1 \frac{1+t(1-t)}{(1-t(1-t))^2} \mathrm{d} t
$$
The latter integral is evaluated integrating by parts and reducing it to the table integral:
$$
  \int_0^1 \frac{1+t(1-t)}{(1-t(1-t))^2} \mathrm{d} t = \left[ \frac{2}{3} \frac{2t-1}{1-t+t^2} + \frac{2}{3 \sqrt{3}} \arctan\left(\frac{2t-1}{\sqrt{3}} \right) \right]_{t=0}^{t=1} =2 \left(\frac{2}{3} + \frac{\pi}{9 \sqrt{3}}\right)
$$
This establishes the result.
Alternatively, you could note that the summand $c_j$ is a hypergeometric term, which means that the ratio of successive terms is a rational function of $j$
$$
   \frac{c_{j+1}}{c_j} = \frac{j+1}{4j+2} = \frac{(j+1)(j+1)}{j+\frac{1}{2}} \frac{1}{4} \frac{1}{j+1}  
$$
which means that the sum corresponds to the defining series of the Gauss's hypergeometric function:
$$\begin{eqnarray}
   \sum_{j=0}^\infty \frac{j! \cdot j!}{(2j)!} &=& \sum_{j=0}^\infty \frac{(1)_j (1)_j}{\left(\frac{1}{2}\right)_j} \frac{(1/4)^j}{j!} = {}_2F_1\left(\left.\begin{array}{cc} 1 & 1\\ & \frac{1}{2} \end{array} \right| \frac{1}{4}\right) \\
 &=& \left.\frac{1}{1-z} \left(1 + \sqrt{\frac{z}{1-z}} \arcsin\sqrt{z}\right)\right|_{z=1/4} = \frac{4}{3} + \frac{2 \pi}{9 \sqrt{3}}
\end{eqnarray}
$$
</s>