# Geometric Series and Triple Integrals

## Homework Statement

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

dx 0 to 1
dy 0 to 1
dz 0 to 1

## The Attempt at a Solution

Not sure how to relate the two of them

## Answers and Replies

Expand the integrand as a geometric series.

Cyosis
Homework Helper
Remember the geometric series, $$\sum_{n=0}^\infty z^n=\frac{1}{1-z}[/itex]. It's pretty important to memorize it. for the Sum of zn = 1/(1-z) let z = xyz so when u would integrate them by dxdydz from 0 to1 on all of them, you would get 1*1*1 or 13 which could be expressed as n3 am i thinking at all on the right track Cyosis Homework Helper you would get 1*1*1 or 13 which could be expressed as n3 You're on the right track, but from that quote I get the idea that your conclusion isn't entirely right. Write down the series you are integrating, then take the sum symbol in front of the integral (which you may do when series converge uniformly). You will quickly notice where the 1/n^3 comes from. [tex]\sum$$$$\int$$xnynzndxdydz =

$$\sum$$$$\int$$1n+1ynzn)/(n+1)dydz =

$$\sum$$$$\int$$1^(2n+2)zn)/(n+1)2dz=

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

do i get or do i get it

Cyosis
Homework Helper
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]