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

Question

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?

Answer 1

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