# Convergence radius of $\sum_{k=0}^\infty \frac{\binom{2k}{k}}{k^k}z^k$

How can one find out the convergence radius of this series?

$$\sum_{k=0}^\infty \frac{\binom{2k}{k}}{k^k}z^k$$

I tried, but I'm confused because of the $$z$$'s and couldn't get a result therefore. WolframAlpha couldn't calculate it either.

I know that one can calculate the convergence radius via

$$r = \frac{1}{\lim \sup_{k \to \infty} \sqrt[k]{|a_k|}}$$

or

$$r = \lim_{k \to \infty} |\frac{a_k}{a_{k+1}}|$$

and to evaluate:

$$|x-x_0| < r \Rightarrow \text{ The power series converges absolutely }$$ $$|x-x_0| > r \Rightarrow \text{ The power series diverges }$$

Can someone show me how it's done here?

$$\lim_{k\to\infty}\frac{(2k+2)!}{(k+1)!^2} \cdot \frac{k!^2}{(2k)!}\cdot\frac{k^k}{(k+1)^{k+1}}\cdot|z| = \lim_{k\to\infty}\frac{4k+2}{(k+1)^2}\cdot\left(1+\frac{1}{k}\right)^{-k}\cdot|z|$$ $$= 0\cdot e^{-1}\cdot|z| = 0$$
Thus the series converges for all $$z$$. The radius of convergence is infinite.