Is there a closed expression for $\sum_{n=0}^\infty \frac{n^2a^n}{n!}$, where $a \in (0,\infty)$?

Question

Is there a closed expression for $\displaystyle\sum_{n=0}^\infty \frac{n^2a^n}{n!}$, where $a \in (0,\infty)$?

I have concluded that the sum converges using the ratio test:

$\displaystyle\frac{(n+1)^2a^{n+1} n!}{n^2a^n(n+1)!}=\frac{(n+1)a}{n^2} \rightarrow 0$ as $n\rightarrow \infty$.

Also, by comparison the following "similar expression" has a closed form:

$$\sum_{n=0}^\infty \frac{na^n}{n!} = a\sum_{n=0}^\infty \frac{na^{n-1}}{n!}=ae^a$$

Answer 1

Start with $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ Differentiate both sides: $$e^x=\sum_{n=0}^\infty\frac{nx^{n-1}}{n!}$$ Multiply by $x$: $$xe^x=\sum_{n=0}^\infty\frac{nx^n}{n!}$$ Differentiate both sides: $$(x+1)e^x=\sum_{n=0}^\infty\frac{n^2x^{n-1}}{n!}$$ Multiply by $x$: $$x(x+1)e^x=\sum_{n=0}^\infty\frac{n^2x^n}{n!}$$

Answer 2

Hint. Note that for $n\geq 2$, $$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}.$$