Find the radius of convergence of this power series: $\sum_{k=0}^{\infty } \binom{2k}{k}x^{k}$

Question

Given: $\sum_{k=0}^{\infty } \binom{2k}{k}x^{k}$

I started by forming it:

$\binom{2k}{k} = \frac{(2k)!}{k!*(2k-k)!} = \frac{(2k)!}{k!*k!}$

Now the problem is, I cannot write $2! * k!$ instead of $(2k)!$, so there doesn't seem to be a way to eliminate one $k!$ in the denominator.

Or shall I start from here with the ratio test?

I can imagine it will end up same or even more complicated because I don't know about that special rule, how to form it better than that (assuming there is a way to do it...).

Answer 1

Hint. As suggested by @Claude Leibovici, one may write, as $k \to \infty$, $$ \left|\frac{\binom{2(k+1)}{k+1}x^{k+1}}{\binom{2k}{k}x^{k}}\right|=\frac{2(k+1)(2k+1)}{(k+1)^2}|x| \to 4|x|. $$ Then the radius of convergence is $R=\dfrac14$.