How to derive the close form of a power series with a ${2n\choose n}$ binomial coefficient? In a step in a proof that the probability to return to origin in a symmetric random walk is $1$ the following combinatorics result seems important:
$$\displaystyle \sum_{n=0}^\infty{2n\choose n}\,x^n = \frac{1}{\sqrt{1-4x}}$$
I understand that the proof involves the generalized binomial theorem, but I don't know how to get about solving the interplay between the $x^n$ in the power series and the binomial coefficients.

Do you know what a taylor series is? Calculate the Taylor expansion of
$$f(x)=(1-4x)^{-1/2},$$
evaluated at $x_0=0$. The taylor series, evaluated at $x_0$, can be obtained by
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
and you should see this summation appearing naturally.
So your main task is to find the derivatives $f^{(n)}(x_0=0)$. Just differentiate $f(x)$ and plug in $x=0$. You should see the pattern. Note that $2n \choose n$$= \frac{2n!}{n!n!}$. Hence $f^{(n)}(x_0=0)$ should be equal to $\frac{2n!}{n!}$.
