Find the convergence radius of $\sum_{n=1}^\infty {\frac{(2n)!}{(n!)^2}}z^n$

Question

I need to find the convergence radius of $$ \sum_{n=1}^\infty {\frac{(2n)!}{(n!)^2}}z^n $$ I proceeded like this: $$ \sum_{n=1}^\infty {\frac{(2n+2)!}{((n+1)!)^2}}z^{n+1} \times {\frac{(n!)^2}{(2n)!}} $$ But I don't see how to proceed further.

Answer 1

$$\sum_{n=0}^\infty {\frac{(2n)!}{(n!)^2}}z^n=\frac{1}{\sqrt{1-4z}}.$$ The nearest point to the origin such that it is not analytic is $z=\frac14$. Therefore the radius of convergence is $\frac14$.

Answer 2

You almost started with the idea but not finished. $$a_n={\frac{(2n)!}{(n!)^2}}\implies \frac{a_n}{a_{n+1}}=\frac{n+1}{4 n+2}\implies R=\frac 14$$

Have a look here in particular paragraph entitled Theoretical radius.