Summing up the series $\sum_{n=0}^\infty C_{2n}^n r^{2n} $ Here $|r|<1/2$, so that the series converge. 
I can do it by using contour inregration. 
$$ S =\sum_n r^{2n} \frac{1}{2\pi i } \oint_C  \frac{1}{z}(z+ \frac{1}{z})^{2n} dz 
 \\
= \frac{1}{2\pi i } \oint_C  \frac{1}{z} r^{2n}  (z+ \frac{1}{z})^{2n}  dz \\
= \frac{1}{2\pi i } \oint_C  \frac{1}{z} \frac{1}{1- r^2 (z+1/z)^2} dz . $$
Here $C$ is the unit circle in the complex plane. 
It is not so tedious to get the final result, which is $1/\sqrt{1-4r^2 }$. 
However, can anyone give a more direct solution? 


We can  use the binomial series expansion. In order to do so we recall the binomial identity
  \begin{align*}
\binom{2n}{n}=(-4)^n\binom{-\frac{1}{2}}{n}\tag{1}
\end{align*}
  and obtain
  \begin{align*}
\color{blue}{\sum_{n=0}^\infty \binom{2n}{n}r^{2n}}&=\sum_{n=0}^\infty (-4)^n\binom{-\frac{1}{2}}{n}r^{2n}\\
&=\sum_{n=0}^\infty\binom{-\frac{1}{2}}{n}(-4r^2)^n\\
&=\left(1-4r^2\right)^{-\frac{1}{2}}\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1-4r^2}}}
\end{align*}

The identity   (1) is  valid since we have
\begin{align*}
\color{blue}{(-4)^n\binom{-\frac{1}{2}}{n}}&=(-4)^n\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{1}{2}-(n-1)\right)\\
&=\frac{2^n}{n!}(2n-1)!!\\
&=\frac{2^n}{n!}\frac{(2n)!}{(2n)!!}\\
&=\frac{2^n}{n!}\frac{(2n)!}{2^nn!}\\
&=\frac{(2n)!}{n!n!}\\
&\,\,\color{blue}{=\binom{2n}{n}}
\end{align*}
