Results 1 to 4 of 4

Math Help - Comparison test for convergence/divergence

  1. #1
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    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 \frac{1}{r^{2}} converges given \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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Comparison test for convergence/divergence

    Quote Originally Posted by Shizaru View Post
    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 \frac{1}{r^{2}} converges given \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 \textstyle\sum 1/r^2 from the divergence of \textstyle\sum 1/r.

    But I suspect that you have misread the question. My guess is that you are told to assume that \textstyle\sum 1/r diverges, and that \textstyle\sum 1/r^2 converges, and that you are supposed to use these facts to answer the remainder of the questions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2011
    From
    United States
    Posts
    41

    Re: Comparison test for convergence/divergence

    the first question is prove the \sum_{r=1}^{\infty}\frac{1}{r^{2}} converges using the comparison test and the fact that \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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Comparison test for convergence/divergence

    Quote Originally Posted by Shizaru View Post
    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 \textstyle\sum 1/r^2 converges. One is to use the integral test and the fact that \int^\infty \frac1{x^2}\,dx converges. The other is to use that fact that \frac1{r^2}<\frac1{r(r-1)} for all r>1. Use partial fractions to see that \sum\frac1{r(r-1)} is a telescoping series which converges. Then the comparison test tells you that  \sum\frac1{r^2} also converges.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: October 26th 2010, 07:20 PM
  2. convergence from comparison test
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: December 1st 2009, 08:44 PM
  3. Replies: 3
    Last Post: November 17th 2009, 08:18 PM
  4. convergence and comparison test help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2009, 02:32 AM
  5. Replies: 3
    Last Post: February 1st 2009, 01:53 PM

Search Tags


/mathhelpforum @mathhelpforum