The problem in the book is to determine the radius of convergence for the given power series, and test for convergence at the boundary points.
$\sum_{n=1}^\infty \frac {(n!)^2}{(2n)!} z^n \qquad z\in\mathbb C$
Using the ratio test, it is clear that the radius of convergence is 4, that is, for $|z|<4$ the series converges. My question is how can we determine convergence at the boundary?
My Attempt
Both the root test and ratio test are inconclusive. It would be helpful if I could take the limit of $\tfrac{(n!)^2}{(2n)!}4^n$, but I don't know how to directly compute $\lim_{n\to\infty}\tfrac{(n!)^2}{(2n)!}4^n$.
In a similar problem I was able to use some inequalities related to $n!$ to conclude that the sequence was always greater than some positive constant. My thinking is that, if I can show that $\tfrac{(n!)^2}{(2n)!}4^n>c$, for some positive constant c, then the series cannot converge for any complex $z$ where $|z|=4$ since the limit of the series would not be zero. I tried that approach, and here are the inequalities: $n^n e^{1-n}
Obviously, this isn't going to cut it. I would appreciate a hint.