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

Solution :

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

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:

\[\begin{eqnarray*}

\sum_{k=1}^{\infty}\frac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}

& = & \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!}

\end{eqnarray*}\]

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

Now if we adjust the indices a little, we obtain:

\[\begin{eqnarray*}

\sum_{k=1}^{\infty}\frac{k-1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{(k-1)!}

& = & \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!}

\end{eqnarray*}\]

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

## 9 comments:

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I'd actually cancel the k's first and then re-index immediately; then instead of the substitution k=(k-1) + 1 coming out of nowhere you see a k+1 in the numerator, which is more intuitive to split up.

Also, here's another way to do it:

First cancel the k's and re-index the sum to start at 0, obtaining $\sum_{k=0}^\infty \frac{k+1}{k!}$.

Now note that this is the same as the series $\sum_{k=0}^\infty \frac{k+1}{k!} x^k$ evaluated at $x=1$.

Observe now that the above sum is the derivative of another sum: $\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^\infty \frac{x^{k+1}}{k!}$.

Factor out an x and you're left with $\frac{\mathrm{d}}{\mathrm{d}x} x \sum_{k=0}^\infty \frac{x^{k}}{k!}$. Apply the series definition of $e^x$ and you get a very workable formula: $\frac{\mathrm{d}}{\mathrm{d}x} (xe^x)$. This is just $e^x + xe^x$.

Thus for the original series you have $\sum_{x=1}^\infty \frac{k^2}{k!} = \left(e^x + xe^x\right|_{x=1}$, from which evaluation at 1 gives you e+e=2e.

Very clever! I think your technique is more useful than my solution since many of these summation identities are about recognizing the correct power series representation.

Thanks! I'll add a page below the header covering basics of MathJax tonight.

Thanks again! There is now a message above the comment box indicating the correct control characters.

Testing reply system.

it's LaTeX with a capital T and you can write it with a backslash in the LaTeX non-math environment. Not sure if it works in math mode. Let's see. $\LaTeX$

Thanks again! The typo is corrected.

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