Is the following series $\sum_{n=0}^\infty \frac{n^3}{n!}$ convergent and if so what is the sum? I started by converting the following $\sum_{n=0}^\infty \frac{n^3}{n!}$ into 
$$\sum_{n=1}^\infty \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!}.$$ Did i get it right and where should I go from here ?

Note that by limit comparison test with $\sum \frac 1{n^2}$
$$\frac{\frac{n^3}{n!}}{\frac 1{n^2}}=\frac{n^5}{n!}\to 0$$
therefore the series converges.
For the sum we can use that
$$\sum_{n=0}^\infty \frac{n^3}{n!}=\sum_{n=1}^\infty \frac{n^2}{(n-1)!}=\sum_{n=0}^\infty \frac{n^2-1+1}{(n-1)!}=\sum_{n=2}^\infty \frac{n+1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$
$$= \sum_{n=2}^\infty \frac{n-2+3}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=\sum_{n=3}^\infty \frac{1}{(n-3)!}+3\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=5e$$
