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Sum n^2/n! from 1 to infinity.

  1. Aug 13, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\sum_1^\infty \frac{n^2}{n!} = [/itex]

    3. The attempt at a solution

    Context: practice Math GRE question I don't know how to answer.

    Well, it's bigger than e and converges by the ratio test. Adding up the first 5 or 6 terms suggests that it converges to 2e. That's good enough for a standardized test, but I'd like to know how to handle this series legitimately.

    Thanks in advance for any help.
     
  2. jcsd
  3. Aug 13, 2011 #2
    It does converge to 2e (by wolfram alpha), but I'm not sure how to derive it.
     
  4. Aug 13, 2011 #3

    SammyS

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    Is [itex]\displaystyle \sum_{n=1}^\infty \frac{n}{(n-1)!} [/itex] any easier ?
     
  5. Aug 13, 2011 #4

    Ray Vickson

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    n^2/n! = n/(n-1)!, so your sum = sum_{n=1..inf} n/(n-1)! = sum_{k=0..inf} (k+1)/k!. Look at (d/dx) sum_{k=0..inf} x^(k+1)/k! .

    RGV
     
  6. Aug 13, 2011 #5
    Thank you Sammy and Ray, I have it now. In case anyone is curious:

    [itex]
    \sum_1^\infty \frac{n^2}{n!}

    = \sum_0^\infty \frac{k+1}{k!}

    = lim_{m \to \infty} (p_m + p_{m-1})
    [/itex]
    where [itex] p_k = \sum_0^m 1/k! [/itex] .
    And that limit is 2e.

    Ray, I see that that derivative, evaluated at 1, is equal to our series. But what good does that do?
     
  7. Aug 13, 2011 #6

    dynamicsolo

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    You can look at this in stages:

    What is [itex] \frac{n}{n!} [/itex] equal to?

    Then what is [itex] \sum_1^\infty \frac{n}{n!} [/itex] equal to? (It will help to keep in mind the infinite series for e .)

    Next, what is [itex] \frac{n (n-1)}{n!} [/itex] equal to? (Write it two different ways: by cancelling terms in numerator and denominator and by simply multiplying out the numerator.)

    Then what would [itex] \sum_2^\infty\frac{n (n-1)}{n!} [/itex] equal? (It may help at any of these stages to write out the first few terms of the series.)

    You can now add your result for [itex] \sum_1^\infty \frac{n}{n!} [/itex] to your new result to find [itex] \sum_1^\infty \frac{n^2}{n!} [/itex] .

    (You could also use this iteratively to find a general result for [itex] \sum_1^\infty \frac{n^p}{n!} [/itex] .)
     
    Last edited: Aug 13, 2011
  8. Aug 13, 2011 #7

    SammyS

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    Ray's sum (to be differentiated) is: [itex]\displaystyle \sum_{k=0}^\infty\frac{x^{k+1}}{k!}[/itex]
    [itex]\displaystyle =x\sum_{k=0}^\infty\frac{x^{k}}{k!}[/itex]

    [itex]=xe^x[/itex]
    d/dx(xex), evaluated at x=1 is 2e .
     
  9. Aug 13, 2011 #8
    Thanks a lot everyone, this was really helpful.

    Regarding the question I just asked, which Sammy just answered, thanks. I was asking myself what function
    [itex]\displaystyle \sum_{k=0}^\infty\frac{x^{k+1}}{k!}[/itex]
    was a power series for, but for some reason it wasn't clicking. I guess I'm pretty rusty with this sort of thing.
     
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