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?
 A: 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: 
$S=\frac{8\sqrt3}{9}\int\limits_0^{\frac{\pi}{6}}\cos^2 \theta d\theta+\frac{4\sqrt3}{3}\int\limits_0^{\frac{\pi}{6}}\cos2\theta d\theta-1=\frac{2\sqrt3\pi}{27}+\frac{1}{3}$
A: Using geometric series (and their derivatives), we get
$$
\sum_{n=0}^\infty(2n+1)x^n=\frac{1+x}{(1-x)^2}\tag1
$$
Let $\alpha=\frac{1+i\sqrt3}2$, then

$$
\begin{align}
\sum_{n=0}^\infty\frac{n!^2}{(2n)!}
&=\sum_{n=0}^\infty(2n+1)\int_0^1t^n(1-t)^n\,\mathrm{d}t\tag2\\
&=\int_0^1\frac{1+t(1-t)}{(1-t(1-t))^2}\,\mathrm{d}t\tag3\\[3pt]
&=\int_0^1\frac{1+t-t^2}{\left(1-t+t^2\right)^2}\,\mathrm{d}t\tag4\\
&=\int_0^1\left(-\frac23\frac1{(t-\alpha)^2}-\frac23\frac1{(t-\bar\alpha)^2}+i\frac1{3\sqrt3}\left(\frac1{t-\bar\alpha}-\frac1{t-\alpha}\right)\right)\mathrm{d}t\tag5\\[3pt]
&=\left[\color{#C00}{\frac23\frac1{t-\alpha}}+\color{#090}{\frac23\frac1{t-\bar\alpha}}+i\frac1{3\sqrt3}\big(\color{#00F}{\log(t-\bar\alpha)}\color{#C90}{-\log(t-\alpha)}\big)\right]_0^1\tag6\\[6pt]
&=\color{#C00}{\frac23}+\color{#090}{\frac23}+\color{#00F}{\frac\pi{9\sqrt3}}\color{#C90}{+\frac\pi{9\sqrt3}}\tag7\\[6pt]
&=\frac43+\frac{2\pi}{9\sqrt3}\tag8
\end{align}
$$
Explanation:
$(2)$: Beta Function
$(3)$: apply $(1)$
$(4)$: expand
$(5)$: Partial Fractions
$(6)$: integrate each term
$(7)$: evaluate at limits
$(8)$: combine
Subtracting the $n=0$ term, we get
$$
\sum_{n=1}^\infty\frac{n!^2}{(2n)!}=\frac13+\frac{2\pi}{9\sqrt3}\tag9
$$
A: If you also accept non-senseful alternatives:
$\displaystyle\sum\limits_{n=1}^\infty\frac{n!^2}{(2n)!} = \sum\limits_{n=1}^\infty\frac{1}{(2n)!} \int\limits_0^\infty\frac{t^n}{e^t}dt \int\limits_0^\infty\frac{s^n}{e^s}ds = \int\limits_0^\infty \int\limits_0^\infty \frac{\cosh(\sqrt{ts})-1}{e^{t+s}} dt ds$
$\displaystyle \int\limits_0^\infty \frac{\cosh(\sqrt{ts})-1}{e^s} dt =  \frac{\sqrt{\pi}}{2}e^{t/4}\sqrt{t}~\text{erf}\left(\frac{\sqrt{t}}{2}\right) $
$\displaystyle a>0 :\enspace \int\limits_0^\infty \frac{\sqrt{t}~\text{erf}\left(\frac{\sqrt{t}}{2}\right)}{e^{at}} dt = \frac{\frac{2\sqrt{a}}{4a+1} + \cot^{-1}(2\sqrt{a})}{\sqrt{\pi}a^{3/2}}\enspace$ ; $~$ here: $\enspace\displaystyle a:=\frac{3}{4}$
see e.g. erf(x)
