Some of series involving factorial in the denominator What would be the sum of the series $\dfrac{n^2}{n!}$ ? 
I don't even know where to start with. It's nothing like telescopic. I tried to compare with some known series but that doesn't seems to work. 
 A: Note that $\frac {n^2}{n!}=\frac n{(n-1)!}=\frac 1{(n-2)!}+\frac 1{(n-1)!}$  You didn't say what the lower limit of $n$ is, so you may need some correction at the bottom end.
A: $$
2e=(xe^x)'(1)=\left(\sum_{n=0}^\infty\frac{x^{n+1}}{n!}\right)'(1)=\left(\sum_{n=0}^\infty\frac{(n+1)x^n}{n!}\right)(1)=\sum_{n=0}^\infty\frac{n+1}{n!}=\sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=1}^\infty\frac{n^2}{n!} 
$$
A: Hint: start with the Maclaurin series for $e^x$, and consider first and second derivatives with respect to $x$.
A: With convergence use ration criterion
$$\lim_\infty\frac{a_{n+1}}{a_{n}}$$
A: Consider $$A=\sum_{n=0}^\infty \frac {n^2}{n!}x^n=\sum_{n=0}^\infty \frac {n(n-1)+n)}{n!}x^n=\sum_{n=0}^\infty \frac {n(n-1)}{n!}x^n+\sum_{n=0}^\infty \frac {n}{n!}x^n$$ $$A=x^2\sum_{n=0}^\infty \frac {n(n-1)}{n!}x^{n-2}+x\sum_{n=0}^\infty \frac {n}{n!}x^{n-1}$$ $$A=x^2\left(\sum_{n=0}^\infty \frac {x^n}{n!} \right)''+x\left(\sum_{n=0}^\infty \frac {x^n}{n!} \right)'=(x^2+x)e^x$$ Now, make $x=1$.
A: In general: $$\sum_{n=1}^{+\infty}\frac{P(n)}{(n+a)!}\text{ with }a\in \mathbb{N} \text{ and }P\text{ polynomial of degree }k.$$ Express$$P(n)=A_k\underbrace{(n+a)(n+a-1)\ldots}_{k\text{ factors}}+A_{k-1}\underbrace{(n+a)(n+a-1)\ldots}_{k-1\text{ factors}}$$ $$+\cdots+A_2\underbrace{(n+a)(n+a-1)}_{2\text{ factors}}+A_1\underbrace{(n+a)}_{1\text{ factor}}+A_0.$$ The indetermined coefficients $A_0,A_1,\ldots,A_k$ can be found giving to $n$ the values $-a,$ $-(a-1),$ etc. Decompose $P(n)$ in the mentioned form and use  $$e=\sum_{m=0}^{+\infty}\frac{1}{m!}.$$
