Via experiments with Wolfram alpha, I found the following closed form of the series in the title: $$ g(x) = \sum_{n \ge 1} \frac{x^n}{\binom{2n}{n}} = \frac{x}{4 - x} + \frac{4}{4 - x} \sqrt{\frac{x}{4 - x}} \sin^{-1}\left(\frac{\sqrt{x}}{2}\right) $$ for $0 < x < 4$, and I want to know how to prove this. Here experiments are the following: once you plug $x = p / q$ with some coprime $p, q$ (I choose both $p$, $q$ as primes), and the results give hints for the general formula, which should be the above equation based on my experiments (but only with $x = 3/2, 4/3, 13/11, 17/13, 29/17, 31/17$). I know that $\sin^{-1}(x)$ has a Taylor expansion with $\binom{2n}{n}$ in its coefficients (explicit formula is given here), but it is not clear for me how it would give $1 / \binom{2n}{n}$ after multiplying the rational and square root terms.
Here's a context: the original question from my friend is to find the value of $$ \sum_{n \ge 1} \frac{(n!)^2 3^n}{(2n+1)!} $$ where Wolfram gives a nice answer. Once we define $$ f(x) = \sum_{n \ge 1} \frac{(n!)^2 x^n}{(2n+1)!}, $$ then $$ \frac{d}{dx} (xf(x^2)) = \sum_{n\ge 1} \frac{(n!)^2 x^{2n}}{(2n)!} = g(x^2), $$ so we can compute the series as $$ \frac{1}{\sqrt{3}}\int_{0}^{\sqrt{3}} g(x^2) dx. $$ Maybe there's a shortcut to compute the series, but my main question is about the above closed form of $g(x)$ anyway.
