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
 A: The simple bound $n!>(n/e)^ne$, which can be found by approximating the logarithm of the factorial with an integral, implies $\frac{e^nn!}{n^n}>e$. So the terms in $\sum_{n=1}^\infty\frac{e^nn!}{n^n}$ don't even tend to $0$, and the series diverges. The same is true of $\sum_{n=1}^\infty\frac{(-e)^nn!}{n^n}$.
A: 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.
A: So, there's much simpler solution wihtout using any advanced tools: 
prove with induction that $ \frac{e^n \cdot n!}{n^n} \geq 1 $ for all $ n \in N $.
for n=1 its trivial. assume for $ n\in N $ so :
$ \frac{e^{n}\cdot e\cdot n!\cdot\left(n+1\right)}{\left(n+1\right)^{n}\cdot\left(n+1\right)}=e\cdot\frac{e^{n}\cdot n!}{\left(n+1\right)^{n}}=e\cdot\frac{e^{n}\cdot n!}{n^{n}\left(1+\frac{1}{n}\right)^{n}}=\frac{e}{\left(1+\frac{1}{n}\right)^{n}}\cdot\frac{e^{n}\cdot n!}{n^{n}} $
now use I.H :
$ \frac{e}{\left(1+\frac{1}{n}\right)^{n}}\cdot\frac{e^{n}\cdot n!}{n^{n}}\geq\frac{e}{\left(1+\frac{1}{n}\right)^{n}} $ 
the sequence $ \left(1+\frac{1}{n}\right)^{n} $ is increasing and converges to e. so for every $ n \in N $ we can say that $ \frac{e}{\left(1+\frac{1}{n}\right)^{n}}\geq1 $ (e is bigger) 
and eventually we have $ \sum_{n=1}^{\infty}\frac{e^{n}n!}{n^{n}}\geq\sum_{n=1}^{\infty}1 $
therefore the series diverge due to Convergence tests. also the limit of the sequence  $ \frac{e^n \cdot n!}{n^n} $ isnt 0, then for a=-e the limit wont be 0 also, so the infinite series diverge for both cases
