Calculate sum of series $\sum \frac{n^2}{n!}$ [duplicate]

I have to calculate sum of series $\sum \frac{n^2}{n!}$. I know that $\sum \frac{1}{n!}=e$ but I dont know how can I use that fact here..

Solution 1:

HINT $$\frac{n^2}{n!}=\frac{n^2}{n\cdot (n-1)!}=\frac{n}{(n-1)!}=\frac{n-1+1}{(n-1)!}=\frac{n-1}{(n-1)!}+\frac{1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$$

Solution 2:

Apply the operator $DxD$, where $D=\frac{d}{dx} $, to

$$\sum_{n=0}^{\infty} \frac{x^n} {n!} = e^x $$

and then substitute $x=1$. See similar techniques.

Solution 3:

Hint:

Let $f(x)=\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$. Express $ g(x)=\displaystyle \sum_{n=0}^{\infty} \frac{n^2x^{n}}{n!} $ in terms of $f'(x)$ and $f''(x)$.

Interpret the result using $f(x)=e^x$.