Results 1 to 3 of 3

Math Help - convergence and divergence

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    7

    convergence and divergence

    Consider
    ∞ ∑ (3k+5/k^2-2k)^pk=3
    for each p ∈ R. show this series - converges if p > 1
    - diverges if p < or = to 1

    Hint: Determine the known series whose terms past the second give an approximate match for the terms of this series. This series is suitable (almost) for using the comp[arison test. Seperate comparisons with it, or closely related series, are needed to establish convergence or divergence of the series according to p. You will need to establish inequalities, based on approximations, to apply the comparison tests.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p

    note:

    \frac{1}{k} < \frac{3}{k} < \frac{3}{k} + \frac{5}{k^2} = \frac{3k+5}{k^2} < \frac{3k+5}{k^2-2k}

    so, if p=1
    \sum_{k=3}^{\infty} \frac{1}{k} diverges..

    if 0<p<1,
    \sum_{k=3}^{\infty} \left(\frac{1}{k}\right)^p also diverges by p-series test.

    also, if p\leq0,
    \sum_{k=3}^{\infty} \left(\frac{1}{k}\right)^p = \sum_{k=3}^{\infty} k^{-p} also diverges..

    hence, \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p diverges for p\leq1 by comparison test.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    \frac{3k+5}{k^2-2k} < \frac{3k+5}{-2k} = -\frac{3}{2}-\frac{5}{2k} < -\frac{5}{2k}

    so, \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p < \sum_{k=3}^{\infty} \left(-\frac{5}{2k}\right)^p

    and \sum_{k=3}^{\infty} \left(-\frac{5}{2k}\right)^p converges for p>1.

    therefore \sum_{k=3}^{\infty} \left(\frac{3k+5}{k^2-2k}\right)^p converges for p>1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence/divergence
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 30th 2011, 07:37 AM
  2. Convergence and Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2010, 12:57 AM
  3. Convergence and divergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 11th 2009, 07:36 PM
  4. convergence or divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2009, 05:42 PM
  5. please help with convergence divergence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 2nd 2009, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum