How to determine the convergence of the series $\sum_{n=1}^\infty\frac{e^nn!}{n^n}$? $$\sum_{n=1}^\infty\frac{e^nn!}{n^n}$$
we cant use the ratio test nor the root test, because the limit will be 1. 
the main question was to determine whether or not the series
$$\sum_{n=1}^\infty\frac{a^nn!}{n^n}$$
converge, for every $a\in\mathbb R$. I found that for every $-e < a < e$ the series absoulutly converges. For every $a>e$ or $a<-e$ the series does not converge (also not conditional convergence) so now I have to determine if for $a=e$ or $a=-e$ the series will converge. (hopefully it will converge for $a=e$ and therefor also for $a=-e$. other wise i'll have to find out also for $a=-e$) thanks.
by the way, im not familier with tools like Rabbe's test or Sterling's formula, so solutions using other ideas will be great. thanks

Without Stirling's formula
When the ratio test is inconclusive, we can try Gauss's test. (An extension of the ratio test.)

Let $u_n > 0$ and suppose
  $$
\frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(n^{-r}), \qquad h \in \mathbb R,r>1
$$
  If $h>1$, then $\sum u_n$ converges; if $h \le 1$ then $\sum u_n$ diverges.

In our case,
$$
u_n := \frac{e^n n!}{n^n},
\\
\frac{u_n}{u_{n+1}}= \frac{1}{e}\left(\frac{n+1}{n}\right)^n 
= 1-\frac{1}{2n} +O(n^{-2})
$$
So we apply Gauss's test with $h=-1/2, r=2$ and conclude the series diverges.

Explanation for the approimation.
$$
\lim_{n\to \infty} \frac{1}{e}\left(\frac{n+1}{n}\right)^n
=\frac{1}{e}\;\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \frac{e}{e} = 1
$$
so the first term is $1$.  Next I claim
$$
\left(\frac{1}{e}\left(\frac{n+1}{n}\right)^n - 1\right)n \to -\frac{1}{2}
$$
so the next term is $-\frac{1}{2n}$.  For that computation:
$$
\left(\frac{1}{e}\left(\frac{n+1}{n}\right)^n - 1\right)n =
\frac{(1+\frac{1}{n})^n-e}{e/n}
$$
This is indeterminate of the form $0/0$.  L'Hopital suggests this limit is the same as the limit of
$$
\frac{(1+\frac{1}{n})^n\left[\log(1+\frac{1}{n})-\frac{1}{n(1+1/n)}\right]}{-\frac{e}{n^2}}
= -\frac{(1+\frac{1}{n})^n}{e}\;\cdot\;
\left[n^2\log\left(1+\frac{1}{n}\right)-\frac{n^2}{n+1}\right];
$$
Now $-\frac{(1+\frac{1}{n})^n}{e} \to -1$, and to compute the other factor:
\begin{align*}
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n} - \frac{1}{2n^2} + O(n^{-3})
\\
n^2\log\left(1+\frac{1}{n}\right) &= n - \frac{1}{2} + O(n^{-1})
\\
\frac{n^2}{n+1} &= n - 1 + O(n^{-1})
\\
n^2\log\left(1+\frac{1}{n}\right)- \frac{n^2}{n+1} 
&= \frac{1}{2} + O(n^{-1})
\\
n^2\log\left(1+\frac{1}{n}\right)- \frac{n^2}{n+1} 
&\to \frac{1}{2}
\end{align*}
so that
$$
-\frac{(1+\frac{1}{n})^n}{e}\;\cdot\;
\left[n^2\log\left(1+\frac{1}{n}\right)-\frac{n^2}{n+1}\right]
\to -\frac{1}{2}
$$
as claimed.
