Results 1 to 2 of 2

Math Help - Radius and interval of convergence

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    76

    Radius and interval of convergence

    Hi, I'm trying to do this question, but I'm stuck.

    Find the radius and interval of convergence of the series:

    x^2n/(n(ln(n))^2)

    The summation from 2 to inifinity.


    I applied the ratio test and got that the limit equals abs(1/x^2). Applying the inequality I got abs(1/x^2) < 1.

    I'm stuck here, because shouldn't the radius of convergence be 1? According to the inequality, the series will converge on two intervals: [-inf, -1) and (1, inf].

    Some help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Because \forall n>1 is |\frac{1}{n^{2}\cdot \ln^{2} n}|<1 and the series...

     \sum_{n=2}^{\infty} x^{2n} (1)

    ... converges for |x|<1, also the series...

     \sum_{n=2}^{\infty} \frac{x^{2n}}{n^{2}\cdot \ln^{2} n} (2)

    ... converges for |x|<1. Ther only difference is that the series (2), unlike the (1), converges also for |x|=1 ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 1st 2010, 10:22 PM
  2. radius and interval of convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 7th 2009, 01:44 PM
  3. Radius and interval of convergence?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 30th 2009, 09:08 PM
  4. Radius and Interval of Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 24th 2009, 04:30 PM
  5. Radius/Interval of Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 24th 2009, 04:11 AM

Search Tags


/mathhelpforum @mathhelpforum