Sum of series involving factorials and number e

Question

Consider $\sum \frac{n^2}{n!}$. How can I find the sum? My attemp: it is equal to $\sum \frac{n}{(n-1)!}$, and we know that $\sum \frac{1}{n!}=e$.

Answer 1

It is known that $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$. If you differentiate wrt $x$ on both sides you'd get $$e^x=\sum_{n=0}^\infty \frac{nx^{n-1}}{n!}\iff xe^x=\sum_{n=0}^\infty \frac{nx^n}{n!}$$ If we differentiate once again, we get this $$(x+1)e^x=\sum_{n=0}^\infty \frac{n^2x^{n-1}}{n!}$$ Finally, evaluating the equation at $x=1$, you get that the series you asked for is $$\sum_{n=0}^\infty \frac{n^2}{n!}=\sum_{n=1}^\infty \frac{n^2}{n!}=2e,$$ where we changed the starting index to $1$ because the $n=0$ term vanishes due to the $n^2$ in the summand.