The convergence of $\sum_{n=1}^\infty (-1)^n\left(\frac{n}{e}\right)^n\frac{1}{n!}$ I'm trying to figure out if the $\sum_{n=1}^\infty (-1)^n\left(\frac{n}{e}\right)^n\frac{1}{n!}$ converges or not. I've tried the Leibnitz test for alternating series, but it leads to Stirling's formula and I was wondering if there's any other way so I could avoid using it. I'll be grateful for any idea. 
 A: $$a_n = \frac{n^n e^{-n}}{n!} $$
is a positive and decreasing sequence with limit zero, hence the series is convergent by Leibniz rule.
$$\text{decreasing}:\qquad \frac{a_{n+1}}{a_n} = \frac{1}{e}\left(1+\frac{1}{n}\right)^n<1. $$
$$\text{convergent to zero}:\left\{
\begin{eqnarray*}\log(n!)&=&\sum_{k=1}^{n}\log(k)=n\log n-\sum_{k=1}^{n-1}k\log\left(1+\frac{1}{k}\right)\\&\geq &n\log n-\sum_{k=1}^{n-1}k\left(\frac{1}{k}-\frac{1}{4k^2}\right)\\&\geq &n\log n-n+\frac{1}{4}\log n.\end{eqnarray*}\right.$$
By the Lagrange inversion theorem (see 1 and 2)  we have
$$ -\frac{W(x)}{1+W(x)} = \sum_{n\geq 1}\frac{(-1)^{n}n^{n}}{n!}\,x^n$$
for any $x$ sufficiently close to the origin, with $W(x)$ being Lambert's function, i.e. the inverse function of $x e^x$.
It follows that
$$ \sum_{n\geq 1}\frac{(-1)^n n^n}{e^n n!} = -\frac{W(1/e)}{1+W(1/e)} $$
and by Newton's method the value of the series is approximately $-0.2178117$.
A: Too long for a comment but not a (complete) answer:
Notice that there is a theorem due to Stirling asserting that for big $n$ one has:
$$n! \approx \sqrt{2n\pi} \left(\frac{n}{e}\right)^n$$
So, in particular, for big $n$, the term of your sum is $(-1)^n\frac{1}{\sqrt{2n\pi}}$ which tells us that it will for sure converge (since it is an alternating sum of decreasing and tending to zero values).
A: It is possible to show convergence while avoiding Leibniz', and other, convergence tests. However, I've used a slightly convoluted route. I'm assuming that's what you meant, not avoiding Stirling's approximation. First, combining odd and even terms with $b_n=a_{2n}-a_{2n-1}$, the series equals
$$\begin{aligned}S&=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(\frac{n}{e}\right)^{n}}{n!}
\\
&=\sum_{n=1}^{\infty}\frac{\left(2n\right)^{2n-1}-e\cdot\left(2n-1\right)^{2n-1}}{e^{2n}\cdot\left(2n-1\right)!}
\end{aligned}
$$
The factorial can be bounded with the lower bound of Stirling's approximation, $\sqrt{2\pi}\ n^{n+\frac12}e^{-n} \le n!$
$$\begin{aligned}|S|\leq
\left|\frac{1}{e\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{\left(\frac{2n}{2n-1}\right)^{2n-1}-e}{\left(2n-1\right)^{\frac{1}{2}}}\right|
\end{aligned}
$$
Note the series on the right is negative so its sign is flipped by the modulus from this point. With $\ln n \leq n-1$, we have $\left(\frac{2n}{2n-1}\right)^{2n-1} = e^{-\left(2n-1\right)\ln\left(1-\frac{1}{2n}\right)}\geq e^{1-\frac{1}{2n}}$
$$\begin{aligned}|S|&\leq
\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{e^{\frac{1}{2n}}-1}{e^{\frac{1}{2n}}\left(2n-1\right)^{\frac{1}{2}}}
\\
&\leq
\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{e^{\frac{1}{2n}}-1}{\left(2n-1\right)^{\frac{1}{2}}}
\end{aligned}
$$
As $e^x=\frac{1}{e^{-x}}\leq\frac{1}{1-x}$, 
$$\begin{aligned}|S|&\leq
\frac{1}{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^{\frac{3}{2}}}
\\
&\leq
\frac{1}{\sqrt{2\pi}}\left(\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}-\sum_{n=1}^{\infty}\frac{1}{\left(2n\right)^{\frac{3}{2}}}\right)
\\
&\leq
\frac{1-\frac{1}{2\sqrt{2}}}{\sqrt{2\pi}}\sum_{n=1}^{\infty}{n^{-3/2}}
\end{aligned}
$$
Finally, as the series on the right has a strictly decreasing summand, we have $\sum_{n=1}^{\infty}{n^{-3/2}}\leq 1+\int_2^\infty (t-1)^{-3/2}\ \mathrm{d}t=3$, so $|S|\leq\frac{3}{4\sqrt{\pi}}\left(2\sqrt{2}-1\right)=0.774$ and $S$ is convergent.
