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Geometric Series and Triple Integrals

  • Thread starter joemama69
  • Start date
  • #1
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Homework Statement



[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

Homework Equations





The Attempt at a Solution



Not sure how to relate the two of them
 

Answers and Replies

  • #2
180
4
Expand the integrand as a geometric series.
 
  • #3
Cyosis
Homework Helper
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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.
 
  • #4
399
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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
 
  • #5
Cyosis
Homework Helper
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you would get 1*1*1 or 13 which could be expressed as n3
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.
 
  • #6
399
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[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
 
  • #7
Cyosis
Homework Helper
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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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