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Homework Help: A sum of a series problem

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the exact sum of the series:

    1/1!3 + 1/2!4 + ... + 1/n!(n+2)


    2. Relevant equations



    3. The attempt at a solution

    The only series I know that look like this are e^x and ln(x) (centered at x = 1). But I do not know how to combine them and I'm not even sure if that's what I'm supposed to do.

    e^x = 1 + x + x^2/2! + ... + x^n/n!

    ln(x) = (x-1) -(1/2)(x-1)^2 + (1/3)(x-1)^3 ...

    ln(x) is missing the factorials and has alternating signs. I tried messing around with its input value but to no avail.

    Any help would be appreciated.
     
  2. jcsd
  3. Jun 14, 2010 #2

    Dick

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    What kind of a series do you get if you integrate e^x? Can you find a way to modify that function so you get your series?
     
  4. Jun 14, 2010 #3
    Thanks for the help.

    integral(e^x) = x + x^2/2! + x^3/3! + ... + x^(n+1)/n!(n+1)

    I had tried that before. Certainly the integral of e^x is the function that most resembles the series, but the values in the denominator of the series' terms are increasing linearly and any thing I plug in for the x's increase exponentially. I thought maybe integrating a second time would do it but that gives the series:

    x^(n+1)/n!(n+1)(n+2)

    Is there something to multiply the series by to get rid of the (n+1) for all of the terms? That would once again require an input that could increase linearly. I'm kind of lost.
     
  5. Jun 14, 2010 #4

    Dick

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    How about if you integrate e^x from 0 to 1? And sure that only gives you n+1 in the denominator. How about x*e^x?
     
  6. Jun 14, 2010 #5
    Thanks again.

    I'm not sure what your second sentence means. I'm pretty sure I'll have to stick in a one by integrating from 0 to 1 in the end but I haven't found the right series yet. x*e^x gives me the same thing as the integral of e^x, but with the factorials shifted to the left. I still have the same problem as before; I can't figure out how to make the linearly increasing fractions appear.
     
  7. Jun 14, 2010 #6

    Dick

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    Ok, what do you get if you integrate the series for x*e^x term by term from 0 to 1? I'm pretty sure your series is in there.
     
  8. Jun 14, 2010 #7
    I get the right answer.

    Thanks so much for all your help!
     
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