1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ratio test for math convergence

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data

    show ## \sum \frac{x^{2}}{(1+x^{2})^{n}} ## converges uniformly on R

    2. Relevant equations



    3. The attempt at a solution

    I know by ratio test it is absolutely convergent for all x in R.

    I am guessing you use m-test. However I do not really understand how m-test works. It is the end of the quarter and my professor pretty much trail blazed through this last chapter in order to finish on time.
     
  2. jcsd
  3. Jun 8, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The first step is to find the ##M_n## in the m-test. Find the maximal value of each function in your sum. Use calculus.
     
  4. Jun 8, 2013 #3
    Ok, so I let ## f_n ##(x) = ## \frac{x^{2}}{(1+x^{2})^{n}} ##

    and took the derivative,

    ## f_n ##'(x) = ## \frac{2x}{(x^{2}+1)^{n+1}} * [1 - (n-1)x^{2}] ##

    when i set it equal to zero and solve I get x = ± ## \frac{1}{(n-1)^{\frac{1}{2}}} ##

    Then I plug it back in to the original function,

    ## f_n ## ## \left ( \frac{1}{(n-1)^{\frac{1}{2}}} \right ) ## = ## \frac{1}{(n-1)(1+\frac{1}{n})^{n}} ##

    before I try to figure out if ## M_n ## = ## \frac{1}{(n-1)(1+\frac{1}{n})^{n}} ## is a convergent series please tell me if I am completely wrong already.
     
  5. Jun 9, 2013 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's not what I get, but the difference shouldn't matter much for the convergence or otherwise. It's roughly 1/(n e) either way, which clearly doesn't converge, so you'll need a more sensitive test.
     
  6. Jun 9, 2013 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The series is just a GP. If the limit is S and the sum of the first r terms is Sr, consider the error term, S-Sr.
     
  7. Jun 9, 2013 #6
    First of all the thesis that ##f_n(x)## in uniformly convergen on [itex]\mathbb{R}[/itex] is not to be taken for sure.

    Those are not the only values by which ##f'(x)=0##

    Something went wrong during this step. You should get:
    ##\frac{\frac{1}{1-n}}{\left(1+\frac{1}{1-n}\right)^n}##
    and with some easy manipulation you get something nice.

    another nicer way is to see that
    ##\frac{x^2}{(1+x^2)^n}<\frac{1+x^2}{(1+x^2)^n}=\frac{1}{(1+x^2)^{n-1}}##
     
  8. Jun 9, 2013 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Nicer, but not convergent.
    That won't give uniform convergence.
     
  9. Jun 9, 2013 #8
    Hi haruspex,

    It will converge in all reals but zero.

    That's why I'm saying the thesis is not true.
     
    Last edited: Jun 9, 2013
  10. Jun 9, 2013 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The expression I passed that comment on did not involve x, so I don't understand your answer.
    Just because an attempt to prove uniform convergence fails, you cannot conclude that it is not uniformly convergent. (It is.)
     
  11. Jun 9, 2013 #10
    OK ok.
    I only wanted to give an easy way for uniform convergence for ##x \in [-\infty,0),(0,\infty]##.
    Then one should treat the zero as a special case.
     
  12. Jun 9, 2013 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You're not going to show uniform convergence by that method even on ##(0,\infty]##. You would be able to show it for ##(\delta,\infty]## for any δ>0, but that's not the same.
     
  13. Jun 11, 2013 #12
    decided to go about a different way. I wrote out the nth partial sum

    ## s_{1}=\frac{x^{2}}{1+x^{2}} ##

    ## s_{2}=\frac{x^{2}}{\left ( 1+x^{2} \right )^{2}} ##
    .
    .
    .

    ## s_{n} = \frac{x^{2}}{1+x^{2}} + \frac{x^{2}}{\left ( 1+x^{2} \right )^{2}} + ... + \frac{x^{2}}{\left ( 1+x^{2} \right )^{n}} ## ## = x^{2}\left ( \frac{1}{1+x^{2}} +...+\frac{1}{\left ( 1+x^{2} \right )^{n}} \right ) ##

    notice

    ## \left ( \frac{1}{1+x^{2}} +...+\frac{1}{\left ( 1+x^{2} \right )^{n}} \right ) ##

    is a geometric series

    thus ## s_{n} = \left (x^{2} + 1 \right ) \left [ 1 - \left ( \frac{1}{1+x^{2}} \right )^{n} \right ] ##

    my question: is that uniformly convergent? I am kind of confused on how to show a sequence is uniformly convergent. I know if I am able to show that then it means the series is uniformly convergent.
     
    Last edited: Jun 11, 2013
  14. Jun 11, 2013 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Now that's a good idea! But I don't think you've got the algebra quite right. Try it again. Consider that the behavior at x=0 might be quite different than elsewhere and you might find it's pretty easy to show that it CAN'T be uniformly convergent.
     
  15. Jun 11, 2013 #14
    maybe it's just late, and my brain is fried. I now end up with

    ## \frac{1-\left ( \frac{1}{1 + x^{2}} \right )^{n}}{1 + x^{2}} ##

    doesn't seem right.
     
    Last edited: Jun 11, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Ratio test for math convergence
Loading...