Results 1 to 3 of 3

Math Help - Uniform convergence....

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Uniform convergence....

    Prove that \Sigma_{n=1}^\infty \frac{x^2}{(x^2+1)^n} uniformly convergence at [a,\infty) for a>0, and not uniformly convergence at [-a,a] for a>0.


    Thank you so much!
    Last edited by Also sprach Zarathustra; August 2nd 2010 at 05:53 AM. Reason: k=n ( thank you Prove It)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,402
    Thanks
    1291
    There isn't any n in the expression...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2010
    Posts
    44
    To prove uniform convergence on [a, \infty) you could use the Weierstrass M-test - Wikipedia, the free encyclopedia.

    You can use the estimation
    \displaystyle\frac{x^2}{(x^2+1)^n}\leq \frac{x^2}{{n\choose 2} x^4}=\frac{1}{x^2} \frac{1}{{n\choose 2}}\leq \frac{1}{a^2} \frac{1}{{n\choose 2}} for n>1

    Hope that helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Uniform convergence vs pointwise convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 15th 2012, 11:03 PM
  2. Replies: 1
    Last Post: October 31st 2010, 07:09 PM
  3. Pointwise convergence to uniform convergence
    Posted in the Calculus Forum
    Replies: 13
    Last Post: November 29th 2009, 08:25 AM
  4. Pointwise Convergence vs. Uniform Convergence
    Posted in the Calculus Forum
    Replies: 8
    Last Post: October 31st 2007, 05:47 PM
  5. Uniform Continuous and Uniform Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2007, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum