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Geometric Series and Triple Integrals

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] 1/(1-xyz)dxdydz = [tex]\sum[/tex]1/n3 from n = 1 to infiniti

    dx 0 to 1
    dy 0 to 1
    dz 0 to 1
    2. Relevant equations



    3. The attempt at a solution

    Not sure how to relate the two of them
     
  2. jcsd
  3. May 14, 2009 #2
    Expand the integrand as a geometric series.
     
  4. May 14, 2009 #3

    Cyosis

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    Remember the geometric series, [tex]\sum_{n=0}^\infty z^n=\frac{1}{1-z}[/itex]. It's pretty important to memorize it.
     
  5. May 15, 2009 #4
    for the Sum of zn = 1/(1-z)

    let z = xyz

    so when u would integrate them by dxdydz from 0 to1 on all of them, you would get 1*1*1 or 13 which could be expressed as n3

    am i thinking at all on the right track
     
  6. May 15, 2009 #5

    Cyosis

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    You're on the right track, but from that quote I get the idea that your conclusion isn't entirely right. Write down the series you are integrating, then take the sum symbol in front of the integral (which you may do when series converge uniformly). You will quickly notice where the 1/n^3 comes from.
     
  7. May 15, 2009 #6
    [tex]\sum[/tex][tex]\int[/tex]xnynzndxdydz =

    [tex]\sum[/tex][tex]\int[/tex]1n+1ynzn)/(n+1)dydz =

    [tex]\sum[/tex][tex]\int[/tex]1^(2n+2)zn)/(n+1)2dz=

    [tex]\sum[/tex]1^(3n+3)/(n+1)3= 1/13 + 1/23 + 1/33... 1/n3

    do i get or do i get it
     
  8. May 15, 2009 #7

    Cyosis

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    It's pretty much right, but if you want to do it a little bit more accurate you should pay attention to the summation indices.

    [tex]\sum_{n=0}^\infty \frac{1}{(n+1)^3}=\sum_{n=1}^\infty \frac{1}{n^3}[/itex]
     
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