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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?

Luchia
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4 Answers4

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$\displaystyle\sum_{x=0}^{\infty}\frac{x^2(1/2)^xe^{-1/2}}{x!}= e^{-1/2}\sum_{x=1}^{\infty}\frac{x(1/2)^x}{(x-1)!}= e^{-1/2}\sum_{x=0}^{\infty}\frac{(x+1)(1/2)^{x+1}}{x!}\\= \displaystyle \frac{1}{2}e^{-1/2}\left(\sum_{x=0}^{\infty}\frac{x(1/2)^x}{x!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)=\frac{1}{2}e^{-1/2}\left(\sum_{x=1}^{\infty}\frac{(1/2)^x}{(x-1)!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right) =\frac{1}{2}e^{-1/2}\left(\frac{1}{2}\sum_{x=0}^{\infty}\frac{(1/2)^x}{x!}+\sum_{x=0}^\infty\frac{(1/2)^x}{x!}\right)=\frac{1}{2}e^{-1/2}\left(\frac{3}{2}e^{1/2}\right)=\frac{3}{4}$

user85798
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\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}

HorizonsMaths
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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: $$\exp(k)=\sum_{x=0}^{+\infty}\dfrac{k^x}{x!}.$$

Michael Hardy
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Sergio Parreiras
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  • I changed \mathrm{exp} to \exp. This not only prevents italicization but also provides proper spacing in expressiosn like $a\exp b$. It is standard. – Michael Hardy Dec 07 '13 at 20:44
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$$ \frac{x^2}{x!} = \frac{x(x-1)}{x!} + \frac{x}{x!} = \frac 1 {(x-2)!} + \frac 1 {(x-1)!} $$

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.

Michael Hardy
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