How do I prove whether the series converges when $a_n = \frac {n! e^n}{n^n}$? The series $\sum a_n$ is the series I get when analyzing the endpoint of a power series, so the ratio test is out of the question. How else can I find whether this particular series converges?
 A: What's below is overkill, as pointed out in the comments. Indeed, once it's established that 
$$
\frac{a_{n+1}}{a_n}
= \frac{e}{(1+\frac{1}{n})^{n}} > 1
$$
then it's clear that the sequence $(a_n)_n$ is inctreasing. Therefore, $a_n \not\to 0$, and the series $\sum_n a_n$ trivially diverges by the limit test.

Since Stirling's approximation is not allowed*, let's try extensions of the ratio test.
Since
$$
\frac{a_{n+1}}{a_n} = \frac{(n+1)e\cdot  n^n}{(n+1)^{n+1}}
= \frac{e\cdot  n^n}{(n+1)^{n}}
= \frac{e}{(1+\frac{1}{n})^{n}}
$$
we have
$$\begin{align*}
n\left( \frac{a_n}{a_{n+1}} -1 \right)
&= e^{-1}n\left( \left(1+\frac{1}{n}\right)^{n} - e\right)
= e^{-1}n\left( e^{n \ln\left(1+\frac{1}{n}\right)} - e\right)\\
&= e^{-1}n\left( e^{n(\frac{1}{n} - \frac{1}{2n^2}+o\left(\frac{1}{n^2}\right))} - e\right)\\
&= e^{-1}n\left( e^{1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)} - e\right)
= n\left( e^{-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)} - 1\right)\\
&= n\left( -\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right)\\
&\xrightarrow[n\to\infty]{} \boxed{-\frac12} < 1
\end{align*}$$
so the series diverges by Raabe's test.

${}^\ast$ Of course, if you want to use it: $n!\displaystyle\operatorname*{\sim}_{n\to\infty} {\sqrt{2\pi n}}\frac{n^n}{e^n}$, so
$$
a_n \operatorname*{\sim}_{n\to\infty} {\sqrt{2\pi n}}
$$
and the series $\sum_n a_n$ (of positive terms) diverges by the limit test.
