# Geometric Series and Triple Integrals

• joemama69
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.

## 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

Expand the integrand as a geometric series.

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

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]