How to find the radius of convergence of $\sum_{n=0}^\infty \frac{n!x^n}{n^n}$? My professor gave me this function, and normally  I would be able to find the ratio, but the $n^n$ is stumping me. How do you find the radius of convergence for this series?
$$
F(x)= \sum_{n=0}^\infty \frac{n!x^n}{n^n}.
$$
 A: You could use the ratio test. The ratio
$$\frac{a_n}{a_{n+1}}=\frac{n!(n+1)^{n+1}}{(n+1)!n^n}
=\frac{(n+1)^{n}}{n^n}=\left(1+\frac1n\right)^n.$$
The limit of this is well-known...
A: $$u_n=\frac{n!x^n}{n^n}\quad\text{and}\quad u_{n+1}=\frac{(n+1)!x^{n+1}}{(n+1)^{n+1}}=\frac{(n+1)n!x^nx}{(n+1)^n(n+1)}.$$ Thus,
$$\begin{align}
\lim_{n\to\infty}\bigg|\frac{u_{n+1}}{u_n}\bigg|&=\lim_{n\to\infty}\bigg|\frac{(n+1)n!x^nx}{(n+1)^n(n+1)}\cdot\frac{n^n}{n!x^n}\bigg|\\
&=\lim_{n\to\infty}|x|\bigg(\frac{n}{n+1}\bigg)^n\\
&=\lim_{n\to\infty}|x|\bigg(\frac{1}{1+\frac{1}{n}}\bigg)^n\quad\text{note that $\big(1+\frac{1}{n}\big)^n\to e$ as $n\to\infty$}   \\
&=\frac{|x|}{e}\end{align}$$
The series converges if $\frac{|x|}{e}<1$, that is, if $|x|<e$, that is, $-e<x<e$. The radius of convergence is $2e$.
A: If $n! \sim \sqrt{2\pi n}\cdot (\frac{n}{e})^n$, then 
$$\sum_{n=1}^\infty \frac{n!x^n}{n^n} \sim \sum_{n=1}^\infty \frac{\sqrt{2\pi n}\cdot (\frac{n}{e})^n \cdot x^n}{n^n} = \sum_{n=1}^\infty \sqrt{2\pi n}\cdot (\frac{x}{e})^n.$$
Hence: $|\frac{x}{e}|<1 \Rightarrow -e<x<e.$
