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Saturday, 6 August 2011

Math GRE - #5

Determine the value of $\sum_{k=1}^{\infty}\frac{k^2}{k!}.$

  • $e$
  • $2e$
  • $(e+1)(e-1)$
  • $e^2$
  • $\infty$

Solution :

Note that \[\sum_{k=1}^{\infty}\frac{k^{2}}{k!}=\sum_{k=1}^{\infty}\frac{k}{(k-1)!} =\sum_{k=1}^{\infty}\frac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}\]

Now if we adjust the indices a little, we obtain:
& = & \sum_{k=0}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\
& = & \sum_{k=1}^{\infty}\frac{k}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\
& = & \sum_{k=1}^{\infty}\frac{1}{(k-1)!}+\sum_{k=0}^{\infty}\frac{1}{k!} \\
& = & \sum_{k=0}^{\infty}\frac{1}{k!}+\sum_{k=0}^{\infty}\frac{1}{k!}

What we have left is just two series definition of $e$ added to each other, hence the answer is $2e$.


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