Does $\sum \frac{n^n}{n!e^n}$ converge or diverge. I have plugged the sum into mathematica and know that the sum diverges but I cannot prove it. When I tried the root test I was able to simplify it to $\lim_{n\to\infty}\frac{n}{(n!)^{\frac{1}{n}}e}$ which I also had to plug into mathematica and got 1 which means the test is inconclusive. The same issue arose when I tried the root test. The only thing left that I can think to do is the comparison test which would require finding something smaller than $\frac{n^n}{n!e^n}$ that diverges but I haven't had any luck.
 A: Lemma: $\left\{\left(1+\frac{1}{n}\right)^{n+1/2}\right\}_{n\geq 1}$ is a decreasing sequence converging to $e$.
Consequence: since we have $n=\prod_\limits{k=1}^{n-1}\left(1+\frac{1}{k}\right)$ for any $n\geq 2$,
$$ n! = \frac{n^{n}}{\prod_\limits{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}=\frac{n^{n+1/2}}{\prod_\limits{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k+1/2}}\leq\frac{n^{n+1/2}}{e^{n-1}} $$
and:
$$ \frac{n^n}{n! e^n} \geq \frac{1}{e\sqrt{n}} $$
trivially implying that $\sum_\limits{n\geq 1}\frac{n^n}{n! e^n}$ is divergent.
A: Since $\displaystyle\lim_{n\to\infty}\frac{n^n}{n!e^n}\div\frac{1}{n}=\lim_{n\to\infty}\frac{n^{n+1}}{n!e^n}=\infty\;$ (as shown below) and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, 
the series $\displaystyle\sum_{n=1}^{\infty}\frac{n^n}{n!e^n}$ diverges by the Limit Comparison Test.

Since $f(x)=\ln x$ is concave down, the Trapezoidal rule estimate gives that
$\displaystyle\int_1^n\ln x \,dx>\ln n!-\frac{1}{2}\ln n,\;$ so $\;\displaystyle\ln n!<n\ln n-n+1+\frac{1}{2}\ln n$ and therefore
$\displaystyle\ln\left(\frac{n^{n+1}}{n!e^n}\right)=(n+1)\ln n-\ln n!-n>(n+1)\ln n-[n\ln n-n+1+\frac{1}{2}\ln n]-n=\frac{1}{2}\ln n-1$.
Then $\displaystyle\lim_{n\to\infty}\ln\left(\frac{n^{n+1}}{n!e^n}\right)=\infty,\;$so $\displaystyle\lim_{n\to\infty}\frac{n^{n+1}}{n!e^n}=\infty$.
