Find the sum: $$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$$
My try:
I played a bit with the coefficient to make it look easier/familiar:
First attempt: $$\begin{align} \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n &= \sum_{n=0}^\infty \frac{n!}{2^n(2n-1)!!}x^n \\ &= \sum_{n=0}^\infty \frac{n!}{(2n-1)!!}\left(\frac x2\right)^n \end{align}$$ Second attempt: $$\begin{align} \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n &= \sum_{n=0}^\infty \frac{n!\cdot n!}{(n+n)!}x^n \\ &= \sum_{n=0}^\infty \frac{1}{{2n \choose n}}x^n \end{align}$$ However, I could not proceed with any of them. Also, I have figured out that the convergence radius is $4$.
My research:
I have also found the same sum has been discussed at AoPS, which unfortunately uses Beta function that my course has not covered yet.
Entering the sum to Wolphram Alpha, I got the following output for the partial sum:
$$\sum_{n=0}^k\frac{(n!)^2x^n}{(2n)!}=\frac{4\sqrt{x}\left(\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)-\frac{2^{2k}k!(k+1)!B_\frac{x}{4}\left(k+\frac{1}{2},\frac{3}{2}\right)}{(2k)!}\right)}{(4-x)^{3/2}}+\frac{4}{4-x}.$$
My background:
As I have already mentioned, I cannot use Gamma, Beta or similar functions. I only know about the convergence theorems on functional series and operations on them. So, I'm looking for some method that uses quite elementary tricks.
Thanks in advance.