$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{2^{2n}}$

Question

I want to examine the convergence of the series $$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{2^{2n}}$$

In case it converges I want to evaluate it.

I tried the D' Alembert theorem but it was inconclusive.I have a feeling it converges but with the use of complex analysis I get :

I used the known fact that $\displaystyle 2\pi i\binom{2n}{n}=\oint_{\gamma}\frac{\left ( 1+z \right )^{2n}}{z^{n+1}}\,{\rm d}z$ whereas $\gamma$ is any closed circle around the origin and then I proceeded as follows:

$$\sum_{n=0}^{\infty}\frac{\left ( 1+z \right )^{2n}}{z^{n+1}}\frac{1}{2^{2n}}=\sum_{n=0}^{\infty}\frac{1}{z}\left ( \frac{(1+z)^2}{4z} \right )^n=\cdots=-\frac{4}{\left ( z-1 \right )^2}$$

because it is a geometric series. Problem is that I cannot apply contour integration around the unit circle because $1$ in on the circle not inside.

So, this leads me to believe that the series divirges as W|A gives me. But I cannot prove that either since the ratio test is incoclusive. So the best approach would be by the comparison test. But compare it to what series? I cannot see an obvious one. In case it converges , around which circle should I apply contour integration to finish things off?

Answer 1

Use Stirling's

$$\frac{(2n)!}{(n!)^2}\frac{1}{2^{2n}}\sim\frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{n^{2n}e^{-2n}2\pi n}\frac{1}{2^{2n}}=\frac{1}{\sqrt{\pi n}}$$

Answer 2

The best approach is by using Raabe's Theorem for a positive series.
Let $$\lambda(n)=n(\frac{x_n}{x_{n+1}}-1)=\frac{n}{2n+1}$$ Since $$\lim_{n\to\infty}\lambda(n)=\frac12<1$$ According to Raabe's Theorem, the series diverges.