Show that this series
$$\sum_{n=1}^{\infty}\dfrac{e^n\cdot n!}{n^n}$$
is divergent.
My try: since
$$u_{n}=\dfrac{e^n\cdot n!}{n^n},\Longrightarrow \dfrac{u_{n+1}}{u_{n}}=\dfrac{e^{n+1}\cdot(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{e^n\cdot n!}=\dfrac{e}{\left(1+\dfrac{1}{n}\right)^n}$$
then
$$\lim_{n\to\infty}\dfrac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\dfrac{e}{\left(1+\dfrac{1}{n}\right)^n}=1$$
so this limit is $1$, therefore I can't prove it by using the Ratio test.
From sos440 suggestion: by using Stirling's approximation, we have $$\dfrac{e^n\cdot n!}{n^n}\approx \sqrt{2n\pi}\to\infty.$$
Maybe this problem requires other methods.