can $\sum_{k=2}^{n} \frac{1}{k!} \sum_{i=0}^{n-k} \frac{(-1)^i}{i!}$ be simplified?

Question

I was working on a probability theory problem and got the following probability as an answer :

$$\mathbb{P}(A) = \sum_{k=2}^{n} \frac{1}{k!} \sum_{i=0}^{n-k} \frac{(-1)^i}{i!}$$

can this sum be simplified to a nicer looking thing ?

Answer 1

One can reduce to a single summation. Let it be $a_n$. Then \begin{align*} \sum_{n=2}^{\infty}a_n x^n&=\sum_{n=2}^{\infty}x^n\sum_{k=2}^{n}\frac{1}{k!}\sum_{j=0}^{n-k}\frac{(-1)^j}{j!} \\&=\sum_{k=2}^{\infty}\frac{1}{k!}\sum_{n=k}^{\infty}x^n\sum_{j=0}^{n-k}\frac{(-1)^j}{j!} \\&=\sum_{k=2}^{\infty}\frac{1}{k!}\sum_{n=0}^{\infty}x^{n+k}\sum_{j=0}^{n}\frac{(-1)^j}{j!} \\&=\left(\sum_{k=2}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{n=0}^{\infty}x^n\sum_{j=0}^{n}\frac{(-1)^j}{j!}\right) \\&=(e^x-1-x)\sum_{j=0}^{\infty}\frac{(-1)^j}{j!}\sum_{n=j}^{\infty}x^n \\&=\frac{e^x-1-x}{1-x}\sum_{j=0}^{\infty}\frac{(-x)^j}{j!}=\frac{1-(1+x)e^{-x}}{1-x}, \end{align*} and $1-(1+x)e^{-x}=\displaystyle\sum_{k=2}^{\infty}(-1)^k\frac{k-1}{k!}x^k$ implies $\color{blue}{a_n=\displaystyle\sum_{k=2}^{n}(-1)^k\frac{k-1}{k!}}$.

In terms of subfactorials, $a_n=1-\dfrac{!n}{n!}-\dfrac{!(n-1)}{(n-1)!}$.