Prove that the series $\sum_{n=1}^\infty \frac{e^nn!}{n^n}$ diverges Prove that the series diverges: $$\sum_{n=1}^\infty \frac{e^nn!}{n^n}$$
I've tried using the ratio test but after all of my calculations I got that $\lim_{x\to\infty}\frac{a_{n+1}}{a_n} = 1$, which I can conclude nothing from.
The root test gave me the same result: $\lim_{x\to\infty}\sqrt[n]{a_n} = 1$
I'm instructed to first try and use Bernoulli's inequality to prove that the sequence $b_n = (1 + \frac{1}{n})^n$ is monotonically increasing.
I tried to follow the instructions, and I got that for each $n\geq1$, $b_n\geq2$. From here, I can conclude that the series $\sum_{n=1}^\infty b_n$ diverges.
from here, I've tried using the limit comparison test for non-negative series and the furthest I got was to the limit: $\lim_{x\to\infty} \frac{e^nn!}{(n+1)^n}$. I can't figure how to simplify that expression any more.
I'm really frustrated as I know I'm missing some minor yet important detail. I feel as though I've done anything I know and implemented any sort of test I've learned until now, with no results. Also, I can't really understand the direction the textbook wanted me to go as for using Bernoulli's inequality. Also, I am not allowed to use Sterling's approximation as I haven't learned nor proved it.
Any sort of help is greatly appreciated!
 A: If you can use Stirling's approximation, you can easily see that $a_n \to \infty$ as shown in the other (now removed) answer, implying divergence of the series.
Without Stirling, simple way is to notice
$$
\frac{a_{n+1}}{a_n} = \frac{e n^n}{(n+1)^n}>1
$$
if and only if 
$$
\left(1+\frac{1}{n}\right)^{n}<e,
$$
but the latter is a well known, since the left side is monotically increasing (you are instructed to show it using Bernoulli inequality) and has limit $e$. Thus $a_{n+1}>a_{n}$ and the sequence is monotically increasing, thus together with $a_1=e$, necessary condition $a_n \to 0$ is not satisfied. So the corresponding series diverges.
A: Apply the logarithm to the $n$th term in the series to  get
$$\tag 1 n + \sum_{k=1}^{n}\ln k - n\ln n.$$
Now $\sum_{k=1}^{n}\ln k \ge \int_1^n \ln x\, dx.$ That last integral equals
$$x\ln x - x \,\big|_1^n = n\ln n -n +1.$$
Thus $(1) \ge 1.$ Exponentiating back shows the $n$th term of the series is $\ge e.$ Thus the $n$th term does not go to $0,$ hence the series diverges.
