Compute $\sum_{n=0}^\infty \frac{(n+1)^2}{n!}$ This is what I've done so far:
$$\sum_{n=0}^\infty\frac{(n+1)^2}{n!} = 1 +4 +4.5 + \frac 83 + \frac{25}{24} +\frac{3}{10} + \frac {49}{720} +\dots$$
I know I need to manipulate $\frac{(n+1)^2}{n!}=\frac{(n+1)(n+1)}{n(n-1)(n-2)}=$

Using the standard Maclaurin series for $e^x$, we find that
$$xe^x=\sum_0^\infty 
\frac{x^{n+1}}{n!}.$$
Differentiating we get
$$xe^x+e^x=\sum_0^\infty (n+1)\frac{x^n}{n!}.$$
Multiply by $x$, and differentiate again. We get
$$x^2e^x+3xe^x+e^x=\sum_0^\infty (n+1)^2\frac{x^n}{n!}.$$
Finally, set $x=1$.
