Convergent or divergent $\sum_{n=1}^{\infty} \frac{e^nn!}{n^n}$?

Question

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.

Answer 1

It looks like the sequence is decreasing: try $\displaystyle\frac{a_n}{a_{n+1}}$.

Answer 2

Hint: you can use the Stirling approximation

Answer 3

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.

Answer 4

$$\ln \frac{n!e^n}{n^n}=\ln n! +n-n\ln n$$ 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 $n\to\infty.$