TITLE: Convergent or divergent $\sum_{n=1}^{\infty} \frac{e^nn!}{n^n}$?
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: $$\ln \frac{n!e^n}{n^n}=\ln n! +n-n\ln n$$ and $$\ln n!=\sum_{k=2}^n\ln k>\sum_{k=2}^n\int_{k-1}^k\ln x\mathrm{d}x=\int_1^n\ln x\mathrm{d}x=x\ln x-x|_1^n=n\ln-n+1$$ so that  $$\ln \frac{n!e^n}{n^n}>1$$ and the series diverges, since its $n$th term does not go to $0$ as $n\to\infty.$