How to determine the convergence of $\sum \frac 1 {n!}\left(\frac n e\right)^n$ using Raabe's test The series is given by :

$$ \sum_{n \geq 1} \frac  1 {n!}\left(\frac n e\right)^n$$

I tried to test its convergence by using a variant of Raabe's test as shown in Bartle's "Introduction to Real Analysis" book. 
However, the limit that I got is $-\infty$, which I think fails Raabe's test (?)
I also tried using comparison test, limit comparison test, ratio test, which all failed. 
Could anyone help me determine whether this series converges or not?  
 A: The term test is also inconclusive since $(n/e)^n/n! \to 0$ as $n \to \infty$.  However, you can quickly establish divergence of the series since Stirling's approximation gives $(n/e)^n/n!  \sim C/\sqrt{n}$.
If you are interested, Raabe's test states that given a series $\sum a_n$ with positive terms, if we obtain 
$$\tag{*}\lim_{n\to \infty}\left(n \frac{a_n}{a_{n+1}} - n - 1 \right) = r,$$
then the series converges if $r > 0$ and diverges if $r < 0$. The case where $r = 0$ is inconclusive.
In this case, $a_n = (n/e)^n/n!$, so
$$\frac{a_n}{a_{n+1}} = \frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}\frac{n^n}{n!e^n} = \frac{e}{(1 + 1/n)^n},$$
and
$$n \frac{a_n}{a_{n+1}} - n - 1 = n\left(\frac{e}{(1+1/n)^n}- 1 \right) - 1.$$
Since $(1 + 1/n)^n \to e$, the term in parentheses converges to $0$. Also, this shows why the ratio and root test are inconclusive -- since $a_{n+1}/a_n \to 1$.
It can be shown that
$$\frac{e}{2n +2} < e - (1 + 1/n)^n < \frac{e}{2n+1}.$$
(A proof of this inequality is given in Problems and Theorems in Analysis I by Polya and Szego.)
Hence,
$$\frac{e}{(1+1/n)^n} \frac{n}{2n+2} - 1< n\left(\frac{e}{(1+1/n)^n}- 1 \right) - 1 < \frac{e}{(1+1/n)^n} \frac{n}{2n+1} - 1.$$
Since the limits of the left-hand and right-hand sides are both $-1/2$, it follows by the squeeze theorem that the limit in (*) is $r = -1/2$ and we can conclude that the series diverges.
