$$ \sum^{\infty}_{j=0} \frac{(j!)^2}{(2j)!} = \frac{2 \pi \sqrt{3}}{27}+\frac{4}{3} $$
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.
Trying to give you something simpler...
Euler found following development for $\arcsin(x)$ : $$f(x):=2\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$ with $\displaystyle\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$ a central binomial coefficient (see too $(8)$ there).
Compute the derivative $f'(x)$ to get : $$f'(x):=4\frac{\arcsin(x)}{\sqrt{1-x^2}}=4\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{n\binom{2n}{n}}$$ Multiply by $x$ and compute the derivative again : $$(xf'(x))':=4\frac{\arcsin(x)}{\sqrt{1-x^2}}-4\frac x{x^2-1}+4\frac{x^2\arcsin(x)}{\sqrt{1-x^2}^3}=8\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{\binom{2n}{n}}$$
replace $x$ by $\frac 12$ (so that $2x=1$ at the right) to get (dividing by $8$) : $$\frac{4\sqrt{3}\,\arcsin\left(\frac 12\right)}{9}+\frac 13=\sum_{n=1}^\infty \frac 1{\binom{2n}{n}}$$ Use $\arcsin(1/2)=\frac{\pi}6$ and add the $n=0$ term at the right ($1$) to conclude that : $$\boxed{\displaystyle\sum_{n=0}^\infty \frac 1{\binom{2n}{n}}=\frac{2\pi\sqrt{3}}{27}+\frac 43}$$
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: $$\begin{eqnarray} \sum_{j=0}^\infty \frac{j! \cdot j!}{(2j)!} &=& \sum_{j=0}^\infty \frac{(1)_j (1)_j}{\left(\frac{1}{2}\right)_j} \frac{(1/4)^j}{j!} = {}_2F_1\left(\left.\begin{array}{cc} 1 & 1\\ & \frac{1}{2} \end{array} \right| \frac{1}{4}\right) \\ &=& \left.\frac{1}{1-z} \left(1 + \sqrt{\frac{z}{1-z}} \arcsin\sqrt{z}\right)\right|_{z=1/4} = \frac{4}{3} + \frac{2 \pi}{9 \sqrt{3}} \end{eqnarray} $$
Here's an alternative method I read from somewhere (Can't remember the source):
We compute the generating function $f(x)$ of $a_j=\frac{4^j}{\binom{2j}{j}}$.
Note that $a_0=1, a_1=2$, and $(2j+1)a_{j+1}=2(j+1)a_j$, so $2(j+1)a_{j+1}-a_{j+1}=2ja_j+2a_j$.
Thus $2f'(x)-\frac{f(x)-1}{x}=2xf'(x)+2f(x)$, so $2x(1-x)f'(x)-(1+2x)f(x)+1=0$
Solving the differential equation with the constraint $f(0)=a_0=1, f'(0)=a_1=2$, we get $f(x)=\frac{1}{1-x}(\sqrt{\frac{x}{1-x}}\arctan{\sqrt{\frac{x}{1-x}}}+1)$.
To get the required sum, we now substitute $x=\frac{1}{4}$, giving $\frac{2\pi\sqrt{3}}{27}+\frac{4}{3}$.
Ok, so $$\int_0^1x^k(1-x)^kdx=\dfrac{(k!)^2}{(2k+1)!}$$ For a proof refer to,Good ways to integrate $\int_0^1 x^{k+1} (1-x)^k dx$?
Let $y= 1-x$ $$\sum_{k=0}^{\infty}=\dfrac{(k!)^2}{(2k)!}=\sum_{k=0}^\infty(2k+1)\int_0^1(xy)^kdx=\int_0^1\dfrac{1}{1-xy}dx+\int_0^1\dfrac{2xy}{(1-xy)^2}dx$$ This is by interchanging order of summation and integration and then using formula for the geometric series and a related series. Evaluating these we get the answer.
