Geometric Series and Triple Integrals

[joemama69]

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

 

[phsopher]

Expand the integrand as a geometric series.

 

[Cyosis]

Remember the geometric series, [tex]\sum_{n=0}^\infty z^n=\frac{1}{1-z}[/itex]. It's pretty important to memorize it.

 

[joemama69]

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

 

[Cyosis]

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.

 

[joemama69]

[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

 

[Cyosis]

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]