Evaluating a simple series

Question

I am having a little trouble with simple computation.

What is $\sum_{k=1}^\infty \frac{k^2}{k!}$?

Answer 1

Notice, $$\sum_{k=1}^{\infty}\frac{k^2}{k!}=\sum_{k=1}^{\infty}\frac{k^2}{k(k-1)!}$$ $$=\sum_{k=1}^{\infty}\frac{k}{(k-1)!}$$$$=\sum_{k=1}^{\infty}\frac{(k-1)+1}{(k-1)!}$$$$=\sum_{k=1}^{\infty}\frac{(k-1)}{(k-1)(k-2)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}$$ $$=\sum_{k=1}^{\infty}\frac{1}{(k-2)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}$$$$=\sum_{k=2}^{\infty}\frac{1}{(k-2)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}=e+e=2e$$