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

Hint: $(n+1)^2=n^2+2n+1=n(n-1)+3n+1$
