How to find the series $\sum_{n=1}^{\infty}\frac{[(n-1)!]^2}{(2n)!}(2x)^{2n},$ 
Find this sum
  $$\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n)!}(2x)^{2n}.\qquad (-1\le x\le 1)$$

My idea: let
$$f(x)=\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n)!}(2x)^{2n},$$
then we have
$$f'(x)=2\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n-1)!}(2x)^{2n-1}$$
$$f''(x)=4\sum_{n=1}^{\infty}\dfrac{[(n-1)!]^2}{(2n-2)!}(2x)^{2n-2}$$
and so on. Then I can't solve it. Thank you for your help.
 A: Here's a derivation of sorts.  Let $f(x)$ be the sum of interest.  Then
$$x \frac{d}{dx} \left [x \frac{d}{dx} f(x) \right ] = 4 \sum_{n=1}^{\infty} \frac{(2 x)^{2 n}}{\displaystyle \binom{2 n}{n}}$$
Referencing this solution, we see that
$$x \frac{d}{dx} \left [x \frac{d}{dx} f(x) \right ] = 4 \frac{x \arcsin{x}}{(1-x^2)^{3/2}} + \frac{4}{1-x^2}-4 $$
Divide by $x$ and integrate to get
$$x \frac{d}{dx} f(x) = 4 \int dx \frac{\arcsin{x}}{(1-x^2)^{3/2}} + 4 \int dx \left [\frac1{x (1-x^2)} - \frac1{x} \right ]$$
To evaluate the integrals, we may use a trig sub in the form of $x=\sin{\theta}$; the integrals become
$$4 \int d\theta (\theta \, \sec^2{\theta} + \csc{\theta} \sec{\theta}-\cot{\theta})$$
Integrate the first piece by parts, and this is equal to
$$4 \theta \tan{\theta} + 4 \underbrace{\int d\theta (\csc{\theta}\sec{\theta}-\cot{\theta}-\tan{\theta})}_{\text{Nice how this vanishes, huh?}}= 4 \theta \tan{\theta} +C $$
Setting $C=0$ from $f'(0)=0$, I get
$$f'(x) = 4 \frac{\arcsin{x}}{\sqrt{1-x^2}} $$
and therefore, noting that $f(0)=0$,
$$f(x) = 2 (\arcsin{x})^2$$
