I am trying to calculate the radius of convergence of
$\sum_{n=1}^\infty \frac{n^n}{(n!)^2}z^n$
with $z \in \mathbb{C}$. According to WolframAlpha, both the root and the ratio test are conclusive, indicating convergence. However at the root test, I am stuck at:
(For convenience I drop the limes superior in the following equations, but it should be in front of every expression.)
$\sqrt[n]{\frac{n^n}{(n!)^2}}=\frac{n}{\sqrt[n]{n!}\sqrt[n]{n!}}$
Is there some estimate of $\sqrt[n]{n!}$ I should know of? For the ratio test, I think I get a reasonable solution:
$\frac{\frac{(n+1)^{n+1}}{(n+1)!^2}}{\frac{n^n}{n!^2}}=\frac{(n+1)^{n+1}n!^2}{(n+1)!^2n^n} = \frac{(n+1)^{n-1}}{n^n} = \frac{1}{n}(1+\frac{1}{n})^n = \frac{e}{n}$
which results in a radius of convergence of $\infty$. I should get the same result with the root test, shouldn't I? After all the root test is stronger.
Hopefully someone can point out of how to continue with the root test.
Thanks in advance
ftiaronsem