Sum of infinite squares and factorials: $\frac{1^2}{1!}+ \frac{2^2}{2!}+ \frac{3^2}{3!} + \frac{4^2}{4!} + \dots$

Question

$$\frac{1^2}{1!}+ \frac{2^2}{2!}+ \frac{3^2}{3!} + \frac{4^2}{4!} + \dotsb$$

I wrote it as: $$\lim_{n\to \infty}\sum_{r=1}^n \frac{(r^2)}{r!}.$$

Then I thought of sandwich theorem, it didn't work. Now I am trying to convert it into difference of two consecutive terms but can't. Need hints.

Answer 1

Hint 1: $$\sum_{r=1}^n \frac{(r^2)}{r!}=\sum_{r=1}^n \frac{r}{(r-1)!}=\sum_{r=0}^{n-1} \frac{r+1}{r!}$$

Hint 2: Derivate $$xe^x=\sum_{r=0}^\infty \frac{x^{r+1}}{r!}$$

Answer 2

The $n^{th}$ term of this series, suppose:

$t_n=\dfrac{n^2}{n!}=\dfrac{n(n-1)+n}{n!}=\dfrac{n(n-1)}{n!}+\dfrac{n}{n!}=\dfrac{1}{(n-2)!}+\dfrac{1}{(n-1)!}$

Sum of terms: \begin{align*} S_n&=\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!} +\cdots\infty\\ &=\sum\limits_{n=1}^{\infty}t_n\\ &=\sum\limits_{n=1}^{\infty}\left[\dfrac{1}{(n-2)!}+\dfrac{1}{(n-1)!}\right]\\ &=\sum\limits_{n=1}^{\infty}\dfrac{1}{(n-2)!}+\sum\limits_{n=1}^{\infty}\dfrac{1}{(n-1)!}\\ &=\left(0+\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!} +\cdots\infty\right)+\left(\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!} +\cdots\infty\right)\\ &=e+e\\ &=2e \end{align*}