Value of $\sum\limits_{n= 0}^\infty \frac{n²}{n!}$ 
How to compute the value of $\sum\limits_{n= 0}^\infty \frac{n^2}{n!}$ ?

I started with the ratio test which told me that it converges but I don't know to what value it converges. I realized I only know how to calculate the limit of a power/geometric series.
 A: Note that, for every $n\geqslant2$,
$$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac1{(n-2)!}+\frac1{(n-1)!},$$
...and watch out for the small $n$ cases.
A: Hint:


*

*The exponential generating function for $a_n = 1$ is $\sum_n a_n\frac{z^n}{n!} = e^z$.

*Differentiating both sides by $z$ and then multiplying by $z$ gets you $\sum_n n \frac{z^{n}}{n!} = ze^z$.

*Doing it again will get you $ze^z+z^2e^z = e^z(z+z^2)$.

*To eliminate $z$'s, substitute $z=1$.


I hope this helps $\ddot\smile$
A: \begin{eqnarray}
\sum_{n=0}^\infty\frac{n^2}{n!}&=&\sum_{n=1}^\infty\frac{n^2}{n!}=\sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=0}^\infty\frac{n+1}{n!}
=\sum_{n=0}^\infty\frac{n}{n!}+\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=1}^\infty\frac{n}{n!}+e\\
&=&e+\sum_{n=1}^\infty\frac{1}{(n-1)!}=e+\sum_{n=0}^\infty\frac{1}{n!}=e+e=2e
\end{eqnarray}
A: The idea for $$\frac{n^r+b_{r-1}n^{r-1}+\cdots+b_1\cdot n}{n!},$$ we can set this to $$\frac{n(n-1)\cdots(n-r+1)+a_{r-1}\cdot n(n-1)\cdots(n-r+2)+\cdots+a_2n(n-1)+a_1\cdot n+a_0}{n!}$$
$$\frac1{(n-r)!}+\frac{a_{r-2}}{(n-r+1)!}+\frac{a_1}{(n-2)!}+\frac{a_1}{(n-1)!}+\frac{a_0}{(n)!}$$
where the arbitrary constants $a_is,0\le i\le r-2$ can be found comapring the coefficients of the different powers of $n$ 
Here let $$\frac{n^2}{n!}=\frac{a_0+a_1n+a_2n(n-1)}{n!}=\frac{a_0}{n!}+\frac{a_1}{(n-1)!}+\frac{a_2}{(n-2)!}$$
$$\implies n^2=a_0+n(a_1-a_2)+a_2n^2$$
$$\implies a_2=1,a_1-a_2=0\iff a_1=a_2,a_0=0$$
So, we have $$\sum_{n=0}^{\infty}\frac{n^2}{n!}=\sum_{n=0}^{\infty}\frac1{(n-1)!}+\sum_{n=0}^{\infty}\frac1{n!}=\sum_{n=1}^{\infty}\frac1{(n-1)!}+\sum_{n=0}^{\infty}\frac1{n!}$$ as $\displaystyle\frac1{(-1)!}=0$
Now, observe  that each summand is $e$(exponential 'e')
