Find the sum: $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$ 
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.
 A: Suppose that $\displaystyle f(x) = \sum_{n\geqslant 0} a_n x^n$ with $a_0=1$.  We shall assume that the series has non-zero radius of convergence.  Then,
\begin{align*}
\frac{1}{2}\frac{d}{dx} \Big( x^2 f(x^2)\Big) = \sum_{n \geqslant 0} (n+1)a_n x^{2n+1} \quad \tag{1}\label{A}
\end{align*}
and
\begin{align*}
2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg) 
&= \sum_{n \geqslant 1} 2(2n-1)a_n x^{2n-1} \\
&=\sum_{n \geqslant 0}2(2n+1) a_{n+1} x^{2n+1} \tag{2}\label{B}
\end{align*}
If $f$ now satisfies the differential equation,
\begin{align*}
2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg) = \frac{1}{2} \frac{d}{dx} \Big(x^2 f(x^2)\Big) \tag{3}\label{C}
\end{align*}
we can equate coefficients in the power series \eqref{A} and \eqref{B} to derive,
\begin{align*}
a_{n+1} = \frac{n+1}{2(2n+1)} a_n, \quad a_0 = 1
\end{align*}
which means
\begin{align*}
a_n = \frac{n!}{2^n(2n-1)(2n-3) \cdots 1} = \frac{(n!)^2}{(2n)!}.
\end{align*}
and
\begin{align*}
f(x) = \sum_{n \geqslant 0} \frac{(n!)^2}{(2n)!} x^n.
\end{align*}
Introduce $\phi(x) = (f(x^2)-1)/x$ so that the differential equation \eqref{C} becomes,
\begin{align*}
(4-x^2)\phi' -3x \phi  = 2
\end{align*}
Multiply this by the integrating factor $(4-x^2)^{1/2}$ to obtain,
\begin{align*}
(4-x^2)^{3/2} \phi' - 3x (4-x^2)^{1/2}\phi=2(4-x^2)^{1/2}
\end{align*}
which is the same as
\begin{align*}
\frac{d}{dx} \Big( (4-x^2)^{3/2} \phi \Big) = 2(4-x^2)^{1/2} \tag{4}\label{D}.
\end{align*}
We  note that by construction $\phi(0)=0$.
The right hand side may be integrated using the substitution $x=2\sin\theta$ to give,
\begin{align*}
\int_0^\xi 2(4-x^2)^{1/2} dx 
&= 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta   d\theta \\
&=8 \Big[  \sin\theta\cos\theta \Big]_0^{\sin^{-1}(\xi/2)} + 8\int_0^{\sin^{-1}(\xi/2)} \sin^2\theta d\theta \\
&=8 (\xi/2) (1-\xi^2/4)^{1/2}+8\int_0^{\sin^{-1}(\xi/2)} 1-\cos^2\theta d\theta \\
&= 2 \xi (4-\xi^2)^{1/2} + 8\sin^{-1}(\xi/2) - 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta d\theta 
\end{align*}
from which we see,
\begin{align*}
\int_0^\xi 2(4-x^2)^{1/2} dx = \xi(4-\xi^2)^{1/2} + 4\sin^{-1}(\xi/2).
\end{align*}
Substituting this in \eqref{D}, and recalling $\phi(0)=0$, we get
\begin{align*}
\phi(x)(4-x^2)^{3/2} = x(4-x^2)^{1/2} + 4\sin^{-1}(x/2)
\end{align*}
yielding
\begin{align*}
f(x) &= \sqrt x \phi(\sqrt x) + 1 \\
&=  1 + \frac{x}{4-x}+\frac{4\sqrt x \sin^{-1}(\frac{\sqrt x }{2})}{(4-x)^{3/2}} \\
&=\frac{4}{4-x}\Bigg(1 + \frac{\sqrt{x} \sin^{-1}(\frac{\sqrt{x}}{2})}{\sqrt{4-x}} \Bigg)
\end{align*}
A: More than tricky ! (and done using a CAS).
$$\sum_{n=0}^p \frac{(n!)^2}{(2n)!}x^n=\frac d {dx}\Bigg[\sum_{n=0}^p \frac{(n!)^2}{(n+1)\,(2n)!}x^{n+1}\Bigg]$$
$$\sum_{n=0}^p \frac{(n!)^2}{(n+1)\,(2n)!}x^{n+1}=x \, _3F_2\left(1,1,1;\frac{1}{2},2;\frac{x}{4}\right)+\color{red}{\text{monster}}$$ $$\color{red}{\text{monster}}=\frac{((p+1)!)^2 x^{p+2} \,
   _3F_2\left(1,p+2,p+2;p+\frac{3}{2},p+3;\frac{x}{4}\right)}{(p+2) (2 (p+1))!}$$
$$\frac d {dx}\Big[x \, _3F_2\left(1,1,1;\frac{1}{2},2;\frac{x}{4}\right)\Big]=\frac{4}{4-x}+\frac{4 \sqrt{x} \sin
   ^{-1}\left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{\frac 32}}$$ and
$$\frac d {dx}\big[\color{red}{\text{monster}}\big]=\frac{x^{p+1} \Gamma (p+2)^2 \,
   _2F_1\left(1,p+2;p+\frac{3}{2};\frac{x}{4}\right)}{\Gamma (2 p+3)}$$ which, in the radius of convergence, tends very quickly to $0$.
