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? 
 A: $\displaystyle\sum_{x=0}^{\infty}\frac{x^2(1/2)^xe^{-1/2}}{x!}=
e^{-1/2}\sum_{x=1}^{\infty}\frac{x(1/2)^x}{(x-1)!}=
e^{-1/2}\sum_{x=0}^{\infty}\frac{(x+1)(1/2)^{x+1}}{x!}\\=
\displaystyle \frac{1}{2}e^{-1/2}\left(\sum_{x=0}^{\infty}\frac{x(1/2)^x}{x!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)=\frac{1}{2}e^{-1/2}\left(\sum_{x=1}^{\infty}\frac{(1/2)^x}{(x-1)!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)
=\frac{1}{2}e^{-1/2}\left(\frac{1}{2}\sum_{x=0}^{\infty}\frac{(1/2)^x}{x!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)=\frac{1}{2}e^{-1/2}\left(\frac{3}{2}e^{1/2}\right)=\frac{3}{4}$
A: \begin{eqnarray}
\sum_{k=0}^\infty\frac{k^22^{-k}e^{-1/2}}{k!}&=&e^{-1/2}\sum_{k=0}^\infty\frac{k2^{-k}}{(k-1)!}=e^{-1/2}\sum_{k=1}^\infty\frac{k2^{-k}}{(k-1)!}=e^{-1/2}\sum_{k=0}^\infty\frac{(k+1)2^{-k-1}}{k!}\\
&=&\frac{1}{2\sqrt{e}}\sum_{k=0}^\infty\left(\frac{k}{k!}+\frac{1}{k!}\right)2^{-k}=\frac{1}{2\sqrt{e}}\sum_{k=1}^\infty\frac{k}{k!}2^{-k}+\frac{1}{2\sqrt{e}}\sum_{k=0}^\infty\frac{2^{-k}}{k!}\\
&=&\frac{1}{2\sqrt{e}}\sum_{k=1}^\infty\frac{2^{-k}}{(k-1)!}+\frac{1}{2\sqrt{e}}\sum_{k=0}^\infty\frac{2^{-k}}{k!}\\
&=&\frac{1}{2\sqrt{e}}\sum_{k=0}^\infty\frac{2^{-k-1}}{k!}+\frac{1}{2\sqrt{e}}\sum_{k=0}^\infty\frac{2^{-k}}{k!}=\left(\frac14+\frac12\right)\frac{1}{\sqrt{e}}\sum_{k=0}^\infty\frac{(2^{-1})^k}{k!}\\
&=&\frac{3}{4\sqrt{e}}\exp(2^{-1})=\frac{3}{4}.
\end{eqnarray}
A: Both answers are correct but I think a less complete answer perhaps is better, what you need to know to solve the problem is the definition of the exponential function: $$\exp(k)=\sum_{x=0}^{+\infty}\dfrac{k^x}{x!}.$$
A: $$
\frac{x^2}{x!} = \frac{x(x-1)}{x!} + \frac{x}{x!} = \frac 1 {(x-2)!} + \frac 1 {(x-1)!}
$$
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.
