Summing up the series $\sum_{n=0}^\infty C_{2n}^n r^{2n} $

Question

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?

Answer 1

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*}