Is it possibile to obtain the sum of the following series without using 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:

$S=\frac{8\sqrt3}{9}\int\limits_0^{\frac{\pi}{6}}\cos^2 \theta d\theta+\frac{4\sqrt3}{3}\int\limits_0^{\frac{\pi}{6}}\cos2\theta d\theta-1=\frac{2\sqrt3\pi}{27}+\frac{1}{3}$

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 $$ \sum_{n=1}^\infty\frac{n!^2}{(2n)!}=\frac13+\frac{2\pi}{9\sqrt3}\tag9 $$