# Comparison test for convergence/divergence

• Jan 9th 2012, 05:23 AM
Shizaru
Comparison test for convergence/divergence
Ok so basically we are told about this comparison test which says if one series is greater or equal than another for all values, and converges, so does the second series. and if the first is less and diverges, the greater one diverges.

I want to know how does the comparison test show that $\displaystyle \frac{1}{r^{2}}$ converges given $\displaystyle \frac{1}{r}}$ diverges?

This is the proof we are asked for but I don't see that there's enough information there. given the first is always less than or equal to the second, for all r, however the comparison test says that if the smaller series diverges, so does the bigger which contradicts this? please help, i am confused, this proof is integral to the rest of the questions and without this I can't procede with my work... S:

i dont see that the comparison test actually answers this
• Jan 9th 2012, 07:36 AM
Opalg
Re: Comparison test for convergence/divergence
Quote:

Originally Posted by Shizaru
Ok so basically we are told about this comparison test which says if one series is greater or equal than another for all values, and converges, so does the second series. and if the first is less and diverges, the greater one diverges.

I want to know how does the comparison test show that $\displaystyle \frac{1}{r^{2}}$ converges given $\displaystyle \frac{1}{r}}$ diverges?

This is the proof we are asked for but I don't see that there's enough information there. given the first is always less than or equal to the second, for all r, however the comparison test says that if the smaller series diverges, so does the bigger which contradicts this? please help, i am confused, this proof is integral to the rest of the questions and without this I can't procede with my work... S:

i dont see that the comparison test actually answers this

You are quite right, it is not possible to deduce the convergence of $\displaystyle \textstyle\sum 1/r^2$ from the divergence of $\displaystyle \textstyle\sum 1/r.$

But I suspect that you have misread the question. My guess is that you are told to assume that $\displaystyle \textstyle\sum 1/r$ diverges, and that $\displaystyle \textstyle\sum 1/r^2$ converges, and that you are supposed to use these facts to answer the remainder of the questions.
• Jan 9th 2012, 07:44 AM
Shizaru
Re: Comparison test for convergence/divergence
the first question is prove the $\displaystyle \sum_{r=1}^{\infty}\frac{1}{r^{2}}$ converges using the comparison test and the fact that $\displaystyle \sum_{r=1}^{\infty}\frac{1}{r}$ diverges. :S

if you say that you cant as i suspected then the mistake is the question itself :S
i should go to a better university maybe

but regardless how else would i prove it converges? the r^2 one.?

or is this just a fundamental theorem we must accept, with a proof too challenging. in which case i dont get why they asked
• Jan 9th 2012, 07:55 AM
Opalg
Re: Comparison test for convergence/divergence
Quote:

Originally Posted by Shizaru
is this just a fundamental theorem we must accept, with a proof too challenging. in which case i dont get why they asked

There are two fairly straightforward ways to show that $\displaystyle \textstyle\sum 1/r^2$ converges. One is to use the integral test and the fact that $\displaystyle \int^\infty \frac1{x^2}\,dx$ converges. The other is to use that fact that $\displaystyle \frac1{r^2}<\frac1{r(r-1)}$ for all r>1. Use partial fractions to see that $\displaystyle \sum\frac1{r(r-1)}$ is a telescoping series which converges. Then the comparison test tells you that $\displaystyle \sum\frac1{r^2}$ also converges.