Functional expression for the sum of a certain power series Does anybody recognize the following power series together with a functional expression for the sum:
$$
\sum_{n = 0}^{\infty} \left( \begin{array}{c} 2n \\ n \end{array} \right) x^n
$$
 A: $$ \sum_{n\geq 0}\binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}} $$
for any $x$ such that $|x|<\frac{1}{4}$ follows from the extended binomial theorem.
As an alternative, you may notice that
$$ \frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos(\theta)^{2n}\,d\theta $$
and convert the previous series into something only depending on
$$ \int_{0}^{\pi/2}\frac{d\theta}{1-4x\cos^2(\theta)} $$
that is simple to compute through the substitution $\theta=\arctan u$.
A: I wanted to elaborate on the alternate derivation Jack suggested. I'm sure there's a shorter way, but here it goes.
One can use induction to prove $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta.$$
The $n=0$ case is clear. For the inductive step note that
$$\binom{2n+2}{n+1} = 4\frac{2n+1}{2n+2}\binom {2n}n,\quad \text{and} \quad \int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = \frac{2n+1}{2n+2}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta,$$
where the second is due to the following partial integration:
\begin{align}
\int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta = 0 + \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)\sin^2(\theta)\,d\theta &= \int_{0}^{\pi/2}(2n+1)\cos^{2n}(\theta)(1-\cos^2(\theta))\,d\theta.
\end{align}
These imply that
$$\frac{1}{4^{n+1}}\binom{2n+2}{n+1} = \frac{1}{4^n}\frac{2n+1}{2n+2}\binom {2n}n = \frac{2n+1}{2n+2}\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n+2}(\theta)\,d\theta.$$
Meaning that the inductive proof is complete, so let's use its result.
\begin{align}
\sum_{n\geq 0}\left ( 4^n\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \right )x^n 
&= \tag{1}
\frac 2\pi \int_{0}^{\pi/2} \left (\sum_{n\geq 0} 4^n \cos^{2n}(\theta) x^n\right ) \,d\theta 
\\&=
\frac 2\pi \int_{0}^{\pi/2} \frac{1}{1-4\cos^2(\theta)x} \,d\theta 
\\&= 
\frac 2\pi \int_{\arctan0}^{\arctan \infty} \frac{1}{1-4\cos^2(\theta)x} \,d\theta 
\\&= \tag{2}
\frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\cos^2(\arctan u)x} \,du 
\\&= \tag{3}
\frac 2\pi \int_0^\infty \frac{\frac 1{1+u^2}}{1-4\frac 1{1+u^2}x} \,du 
\\&= 
\frac 2\pi \int_0^\infty \frac 1 {(1-4x)+u^2} \,du 
\\&= 
\frac 2\pi \left[ \frac{\arctan \left( \frac u {\sqrt{1-4x}} \right)
}{\sqrt{1-4x}}\right]_{u=0}^{\infty}
\\&= 
\frac 2\pi \frac{\frac \pi 2}{\sqrt{1-4x}}
\\&= 
\frac 1 {\sqrt{1-4x}}
\end{align}
Further explanation:
(1) 

 Recognize that this is a geometric series, which converges for $|x| < \frac 14$, since $|x| < \frac 14 \Rightarrow |4\cos^2(\theta)x| < 1.$ By analyticity, we may exchange the order of integrations. It also has a nice closed form.

(2)

 Substitute $\theta = \arctan u$, use $\arctan' = \frac 1 {1+\operatorname{id_{\mathbb R}}^2}. $

(3)

 For example, you can use $\tan^2(\theta) = \frac 1 {\cos^2(\theta)} - 1 $.

