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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:
 \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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[hide] Dave said...

Nice.

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.

on 6 August 2011 at 10:08
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[hide] Paul Liu said...

@Dave:

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.

on 6 August 2011 at 13:50
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[hide] Dave said...

A comment about MathJax: you should mention somewhere (sidebar?) how MathJax syntax works. I had to poke around the MathJax site to figure it out.

on 6 August 2011 at 13:56
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[hide] Paul Liu said...

@Dave:

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

on 6 August 2011 at 14:14
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[hide] Dave said...

I think you can get away with just saying what the control characters to enter the math environment are. The rest is just LaTeX, for which there exist lots of resources and even more sample LaTeX code.

on 6 August 2011 at 15:44
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[hide] Paul Liu said...

@Dave:

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

on 6 August 2011 at 17:03
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[hide] Paul Liu said...

Testing reply system.

on 6 August 2011 at 17:37
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[hide] Dave said...

didn't go through, posting again:

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

on 6 August 2011 at 18:09
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[hide] Paul Liu said...

Thanks again! The typo is corrected.

on 6 August 2011 at 18:32
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