What is the closed form of this series: $\sum_{n\geq 1}\frac{n^k{(-1)}^{n+1}}{n!}$ for $k<-10$ and for $k>1$?

Question

I would like to check the closed form of this sum $$\sum_{n\geq 1}\frac{n^k{(-1)}^{n+1}}{n!}$$ , for an integer $k>1$ and $k<-10$.

Note : I run some computation in wolfram alpha i have got :for $k=2$ $$\sum_{n\geq 1}\frac{n^k{(-1)}^{n+1}}{n!}=0$$ and for $k>2$ The sum always gives :$\frac{m}{e}$ where $m$ is integer ,in the same time the sum of series gives $1$ for $k=-11,-12,-13,\cdots $

Thank you for any kind of help

Answer 1

If you first consider that: $$ \sum_{n\geq 1}\frac{(-1)^{n+1} e^{nx}}{n!} = 1-e^{-e^{x}} \tag{1}$$ you get that for $k\geq 0$ $$ \sum_{n\geq 1}\frac{(-1)^{n+1} n^k }{n!} = \frac{d^k}{dx^k}\left.\left(1-e^{-e^{x}}\right)\right|_{x=0}\tag{2}$$ holds. That gives that your constants relate with Bell numbers.

The sum for $k=-11$ does not equal $1$, but something very close to $1$, $\approx 0.999756790446$, that is the value of a $_{12} F_{12} $ hypergeometric function at $-1$.