Is it possibile to obtain the sum of the following series without using hypergeometric functions? I know that the following series:
$$
\sum_{n=1}^{+\infty}\frac{ (n!)^2}{(2n)!}
$$
converges. If I plug it in Wolphram Alpha, I can see that its sum is
$$
\frac{1}{27} \left(9 + 2 \sqrt{3} \pi\right).
$$ 
Is it possibile to obtain it without the use of the hypergeometric functions?

I started from the fact that 
$\Gamma(n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt{\pi}$ 
}
Divide both side by $n!$ and express $\frac{(n!)^2}{(2n)!}$
We get 
$\frac{(n!)^2}{(2n)!}=\frac{n!}{4^n \Gamma(n+\frac{1}{2})}\sqrt\pi=\frac{\Gamma(n+1)}{4^n \Gamma(n+\frac{1}{2})}\Gamma(\frac{1}{2})$
Introduce $\beta$ function and take the sum of both sides from $0$ to $\infty$
$S=\sum\limits_{n=1}^\infty\frac{(n!)^2}{(2n)!}=\sum\limits_{n=0}^\infty \frac{2n+1}{2^{2n+1}}\beta(n+1,\frac{1}{2})-1$
Based on the definition of $\beta$ function can be written
$S=\sum\limits_{n=0}^\infty \frac{2n+1}{2^{2n+1}}\int\limits_0^1\frac{t^n}{ \sqrt{1-t}} dt-1=\int\limits_0^1\sum\limits_{n=0}^\infty\frac{2n+1}{2^{2n+1}}\frac{t^n}{ \sqrt{1-t}}dt-1$
The integral can be devided into two parts: 
$\int\limits_0^1\frac{1}{\sqrt{1-t}}\sum\limits_{n=0}^\infty n(\frac{t}{4})^n dt+\frac{1}{2}\int\limits_0^1\frac{1}{\sqrt{1-t}}\sum\limits_{n=0}^\infty (\frac{t}{4})^n dt-1 $
Take  $\frac{t}{4}\sum\limits_{n=0}^\infty n(\frac{t}{4})^{n-1}=t\frac{d}{dt}\sum\limits_{n=0}^\infty (\frac{t}{4})^n=\frac{4t}{(4-t)^2}$  and 
$\sum\limits_{n=0}^\infty (\frac{t}{4})^n=\frac{4}{4-t}$ into account we get:
$S=\int\limits_0^1\frac{2(4+t)}{\sqrt{1-t}(4-t)^2}dt-1$
Let $x=\sqrt{1-t}$ then $S=4\int\limits_0^1 \frac{(5-x^2)}{(3+x^2)}dx-1$
Forming the integral in the following way:
$S=\frac{8}{9}\int\limits_0^1\frac{1}{(1+(\frac{x}{\sqrt3})^2)^2}dx+\frac{4}{3}\int\limits_0^1\frac{1-(\frac{x}{\sqrt3})^2}{(1+(\frac{x}{\sqrt3})^2)^2}dx-1$
Applying the following substitution:$\frac{x}{\sqrt3}=\tan \theta$ we receie: 
$S=\frac{8\sqrt3}{9}\int\limits_0^{\frac{\pi}{6}}\cos^2 \theta d\theta+\frac{4\sqrt3}{3}\int\limits_0^{\frac{\pi}{6}}\cos2\theta d\theta-1=\frac{2\sqrt3\pi}{27}+\frac{1}{3}$
