Convergent or divergent $\sum_{n=1}^{\infty} \frac{e^nn!}{n^n}$? Any suggestion/hint, not the whole solution, how to determine convergence/divergence of $$ \sum_{n=1}^{\infty}\dfrac{e^n \cdot n!}{n^n} $$ I'm currently stuck.

OK, apologies, I take my words back about Stirling's approximation. In T. Worsch's article 'Lower and upper bounds on the binomial coefficients' it is shown that $n!$ can be lower-bounded by $(\epsilon>0)$
$$
n! > \frac{1}{1+\epsilon} \bigg(\frac{n}{e}\bigg)^n\sqrt{ 2 \pi n}
$$
If you use this lower bound on the sum you have got, you'll immediately get the divergence of the series. 
