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Evaluate this series:

  1. Dec 20, 2003 #1
    Evaluate this series: Sum from k = 1 to k = infinity k^2/k!

    The answer is 2e.

    Thanks so much!!
     
  2. jcsd
  3. Dec 20, 2003 #2

    Hurkyl

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    What have you tried so far?
     
  4. Dec 20, 2003 #3
    I have no idea how to do this problem. Does it require more knowledge than second year college math? I thought it just requires some knowledge about series, but I can't seem to get the answer. It's been over 3 years since I took calculus so I don't remember much about series expansions. I've tried comparing the series to several common expansions but I can't find the right form or find the right starting point. Thanks!
     
  5. Dec 20, 2003 #4
    yxgao,
    here's a hint:
    [tex]
    e = \sum_{k=0}^\infty \frac{1}{k!}
    [/tex]
    Try to express your sum in terms of this...
     
  6. Dec 20, 2003 #5

    Hurkyl

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    Have you ever heard of a trick called "integrating with respect to 1"?


    Anyways, one of the first things I'd do is simplify the summand; something cancels.
     
  7. Dec 20, 2003 #6

    krab

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    Here's a hint: replace the index k by k+1. You'll be able to cancel k+1 in numerator and denominator. The rest is easy.
     
  8. Dec 27, 2003 #7
    How do you simplify (n+1)/n! summed from n = 0 to n = infinity?
     
  9. Dec 27, 2003 #8

    Hurkyl

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    distribute!
     
  10. Dec 27, 2003 #9
    I don't understand how to proceed from this step. What do you do with the (n+1) in the numerator? What is there to distribute? Can someone post a direct method to the solution?
    Thanks.
     
  11. Dec 27, 2003 #10

    Hurkyl

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    [tex]
    \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}
    [/tex]
     
  12. Dec 27, 2003 #11
    Ah, ok. Why didn't I see that before? :)
    e + e = 2e
    Thanks a lot.
     
  13. Dec 27, 2003 #12

    Hurkyl

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    It's a slick trick, I haven't seen this particular sum done this way before. The method I know can be done as follows (it's more complicated, but more powerful):


    The taylor series for [itex]e^x[/itex] is:

    [tex]e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}[/tex]

    I can insert a [itex]k[/itex] term by differentiating:

    [tex]
    e^x = \sum_{k=0}^{\infty} \frac{k x^{k-1}}{k!}
    [/tex]

    I can then multiply by [itex]x[/itex]:

    [tex]x e^x = \sum_{k=0}^{\infty} \frac{k x^{k}}{k!}[/tex]

    and differentiate again

    [tex]
    (1 + x)e^x = \sum_{k=0}^{\infty} \frac{k^2 x^{k-1}}{k!}
    [/tex]

    Then, plug in [itex]x = 1[/itex] to get the sum.
     
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