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.
Calculus – Does Sum of n^n/n!e^n Converge or Diverge?
calculusconvergence-divergencesequences-and-series
Best Answer
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.