What will be Summation $\sum_{n=1}^{\infty}\frac{n^2}{n!}$

Question

I break it down into two parts the numerator one into $\dfrac{n(n+1)(2n+1)}{ 6}$ but what should be summation of $n!$ ?

Answer 1

$$\sum_{n=1}^{\infty}\dfrac{n^2}{n!}=\sum_{n=1}^{\infty}\dfrac{n}{(n-1)!}=\sum_{n=1}^{\infty}\dfrac{n-1+1}{(n-1)!}=\sum_{n=1}^{\infty}\dfrac{n-1}{(n-1)!}+\sum_{n=1}^{\infty}\dfrac{1}{(n-1)!}$$

$$=\sum_{n=1}^{\infty}\dfrac{1}{(n-2)!}+\sum_{n=1}^{\infty}\dfrac{1}{(n-1)!}=e+e=2e$$

Answer 2

Hint: Consider the first two derivatives of the exponential series.