- #1

- 399

- 0

## Homework Statement

[tex]\int[/tex] 1/(1-xyz)dxdydz = [tex]\sum[/tex]1/n

^{3}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

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- Thread starter joemama69
- Start date

In summary, the given conversation discusses the process of integrating an expression using geometric series and the importance of memorizing the series. The result is expressed as a sum of n^3 terms, with the summation index properly adjusted.

- #1

- 399

- 0

[tex]\int[/tex] 1/(1-xyz)dxdydz = [tex]\sum[/tex]1/n

dx 0 to 1

dy 0 to 1

dz 0 to 1

Not sure how to relate the two of them

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- #2

- 180

- 4

Expand the integrand as a geometric series.

- #3

Homework Helper

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- #4

- 399

- 0

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 1

am i thinking at all on the right track

- #5

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

- 0

[tex]\sum[/tex][tex]\int[/tex]1

[tex]\sum[/tex][tex]\int[/tex]1^(2n+2)z

[tex]\sum[/tex]1^(3n+3)/(n+1)

do i get or do i get it

- #7

Homework Helper

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

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