Prove with mathematical induction (or with a combinatorics) that $E(x)=\sum\limits_{n=0}^{\infty}{(D_n x_n)/n!} = e^{-1}/(1-x)-1$?

Question

$E(x) = \sum_{n=0}^{∞} \frac{D_n*x_n}{n!}$

I think $E(x)$ should equal to either $(e^-1)/(1-x)-1$ or $(e^-1)/(1-x)$ in closed form. Either way, I don't know how to prove that $E(x)$ equals to one of these solutions (to the correct one). $D_n$ is the derangement number and it's equal to: $D_n = \sum_{i=0}^{n} \frac{(-1)^i}{i!}$

I approached this by finding the $E(x)$ in closed form (which I believe is going to be either $(e^-1)/(1-x)-1$ or $(e^-1)/(1-x))$. After that, I tried to prove this using mathematical induction or some combinatorics proving method, but I am not sure how.

Could someone explain this please?

Answer 1

Use the identity $D_n=nD_{n-1}+(-1)^n$, which is easily proved by induction on $n$: multiply by $\frac{x^n}{n!}$ and sum over $n$ to get

$$\begin{align*} \sum_{n\ge 0}D_n\frac{x^n}{n!}&=\sum_{n\ge 0}D_{n-1}\frac{x^n}{(n-1)!}+\sum_{n\ge 0}(-1)^n\frac{x^n}{n!}\\ &=x\sum_{n\ge 0}D_n\frac{x^n}{n!}+\sum_{n\ge 0}(-1)^n\frac{x^n}{n!} \end{align*}$$

and solve for $E(x)=\sum_{n\ge 0}D_n\frac{x^n}{n!}$.

You can find a derivation of $E(x)$ starting from the identity $D_{n+1}=nD_n+nD_{n-1}$ here and a proof of that identity here.