For what complex $z$ the series converges Can someone help me with this assignment?

Find for what $z \in \mathbb{C}$ the series converges
  $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2}z^n$. 

I've just calculated (by using Cauchy-Hadamard theorem) that for all $z$ such that $|z|<\frac{1}{4}$ this series converges and for all $z$ such that $|z|>\frac{1}{4}$ it doesn't. I don't know how to check $|z|=\frac{1}{4}$.
 A: First $\frac{1}{\sqrt{1-4z}}$ has the Taylor expansion $\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} z^n$ in the vicinity of $z = 0$.
For $z \ne \frac14$ on the circle $|z| = \frac14$, the function $\frac{1}{\sqrt{1-4z}}$ is regular there and there is no apriori reason why the sequence must diverge. In fact, if one use the Stirling's approximation of $n!$ and look at the series at $z = \frac14 e^{i\theta}$, we have
$$\binom{2n}{n} z^n \sim \frac{1}{\sqrt{\pi n}} e^{in\theta} ( 1 + O(\frac{1}{n} ))\quad
\text{for large }n.$$
Since the correction term has the order $O(n^{-\frac32})$, its "contribution" in the original series always converges. This means the two series
$$\sum_{n=0}^{\infty}\binom{2n}{n} z^n\quad\text{ and }\quad
\sum_{n=0}^{\infty}\frac{1}{\sqrt{\pi n}} e^{in\theta}$$
converges and diverges at the same time. By Dirichlet test, the second series does converge when $\theta \ne 0$. From this, we can conclude the series $\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n} z^n$ converges for all $z \ne \frac14$ on the circle $|z| = \frac14$.
