TITLE: A finite alternating sum
QUESTION [18 upvotes]: We have stumbled upon the following finite alternating sum, which we have trouble analyzing. The sum is:
$$
S_n = \sum_{j=0}^n \frac{ (-1)^j e^{-j} }{j!} (n-j)^j
$$
We have observed numerically that $S_n \approx 2 n e^{-n}$. We would like to establish whether this conjecture is true. More precisely, we would like to show that $S_n= \Theta(n e^{-n})$.
The sum is quite hard to evaluate numerically because the summands grow in absolute value initially and then decrease toward zero. The largest term is exponential in $n$ while the sum is conjectured to converge to zero exponentially fast. Any references or ideas are welcome!

REPLY [18 votes]: WhatsUp's generating function allows to get an asymptotics: we have $$\sum S_n t^n=\frac{t}{e^{t/e}-t},\sum S_n e^nt^n=\frac{t}{e^{t-1}-t}.$$
Denote $t-1=x$, then $$\frac{t}{e^{t-1}-t}=\frac{1+x}{e^x-1-x}=\frac{1+x}{x^2/2+x^3/6+\ldots}=\frac{2(1+x)}{x^2(1+x/3+\ldots)}=\frac2{x^2}+\frac{4/3}{x}+O(1)$$
for small $x$. Thus $$\frac{t}{e^{t-1}-t}=\frac2{(1-t)^2}-\frac{4/3}{1-t}+F(t)=\sum \left(2n+\frac23\right)t^n+F(t),$$
where $F$ does not have pole at 1. I claim that there is also no pole with absolute value at most 1, that is, the function $e^{t-1}-t$ does not have zeroes in the closed unit disc other than 1. Indeed, assume that $0\leqslant |r|\leqslant 1$ and $\varphi\in (-\pi,\pi]$ and $e^{re^{i\varphi}-1}=re^{i\varphi}$. Taking the arguments of both sides we get $r\sin \varphi=\varphi$ that yields $\varphi=0$, otherwise $|\varphi|>|\sin \varphi|\geqslant r|\sin \varphi|$. So $r=e^{r-1}\geqslant 1+(r-1)=r$ with equality only for $r=1$. This all implies that $F$ is analytic in a disc with radius greater than 1, therefore the coefficients of $F$ are $O(a^n)$ for certain $a\in (0,1)$ and in your question we get $$S_n=(2n+2/3)e^{-n}(1+O(a^n)).$$