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?
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.