Simplifying sum with factorial in denominator I am trying to to find the sum of this series: $\sum_{x=0}^{\infty}\frac{x^2(1/2)^xe^{-1/2}}{x!}$, but I am stuck because I don't know how to deal with the factorial in the denominator. Is this perhaps somehow related to a Taylor series expansion?
$\displaystyle\sum_{x=0}^{\infty}\frac{x^2(1/2)^xe^{-1/2}}{x!}=
e^{-1/2}\sum_{x=1}^{\infty}\frac{x(1/2)^x}{(x-1)!}=
e^{-1/2}\sum_{x=0}^{\infty}\frac{(x+1)(1/2)^{x+1}}{x!}\\=
\displaystyle \frac{1}{2}e^{-1/2}\left(\sum_{x=0}^{\infty}\frac{x(1/2)^x}{x!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)=\frac{1}{2}e^{-1/2}\left(\sum_{x=1}^{\infty}\frac{(1/2)^x}{(x-1)!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)
=\frac{1}{2}e^{-1/2}\left(\frac{1}{2}\sum_{x=0}^{\infty}\frac{(1/2)^x}{x!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)=\frac{1}{2}e^{-1/2}\left(\frac{3}{2}e^{1/2}\right)=\frac{3}{4}$
$$
\frac{x^2}{x!} = \frac{x(x-1)}{x!} + \frac{x}{x!} = \frac 1 {(x-2)!} + \frac 1 {(x-1)!}
$$

Next, break the sum into two sums: one with the first of these two fractions and one with the second.  And notice that the cases $x=0$ and $x=1$ require some special attention.