Results 1 to 5 of 5

Math Help - Convergence/Divergence Question

  1. #1
    Junior Member
    Joined
    Aug 2011
    Posts
    43
    Thanks
    1

    Unhappy Convergence/Divergence Question

    I have spent over 5 hours trying to solve this question to no avail.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927

    Re: Convergence/Divergence Question

    What have you tried? What can you say about the inequality 3k + 5 < k^2 - 2k when k \geqslant 3. Once you've established that, what bounds can you put on \frac{3k + 5}{k^2 - 2k}? Hence, how does the value of p affect the partial sums? Use either the comparison test (which I think you are meant to use) or another method to determine the convergence/divergence of the sum in relation to p?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Convergence/Divergence Question

    Quote Originally Posted by ineedhelplz View Post
    I have spent over 5 hours trying to solve this question to no avail.
    Since for large k:

    \frac{3k+5}{k^2-2k} \sim \frac{3}{k}

    if show that there is a natural N_0 such that for all k>N_0, there exist positive constanta A and B such that:

    A \frac{3}{k}< \frac{3k+5}{k^2-2k}< B \frac{3}{k}

    convergence of the given series is reduced to convergence of:

    \sum_{k=3}^{\infty} \frac{1}{k^p}

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2011
    Posts
    43
    Thanks
    1

    Re: Convergence/Divergence Question

    Quote Originally Posted by CaptainBlack View Post
    Since for large k:

    \frac{3k+5}{k^2-2k} \sim \frac{3}{k}

    if show that there is a natural N_0 such that for all k>N_0, there exist positive constanta A and B such that:

    A \frac{3}{k}< \frac{3k+5}{k^2-2k}< B \frac{3}{k}

    convergence of the given series is reduced to convergence of:

    \sum_{k=3}^{\infty} \frac{1}{k^p}

    CB
    Thank you for your reply.
    I'm unsure as to the next step I'm supposed to take. I presume that the first step "Determine the known series whose terms past the second give an approximate match for the terms of this series." is solved by what you wrote here:
    \frac{3k+5}{k^2-2k} \sim \frac{3}{k}

    however, I am unsure how to do the second step "establish inequalities, based on approximations (as below), to apply the comparison test". I know how to apply the comparison test, but don't know how to use the approximations...
    Last edited by ineedhelplz; August 9th 2011 at 04:21 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Convergence/Divergence Question

    Quote Originally Posted by ineedhelplz View Post
    I have spent over 5 hours trying to solve this question to no avail.

    With limit comparison test.

    A is series with a_k=(\frac{3k+5}{k^2-2k})^p , and B is a series with b_k=(\frac{3}{k})^p

    \lim_{k\to\infty} \frac{a_k}{b_k}=\lim_{k\to\infty}(\frac{\frac{3k+5  }{k^2-2k}}{\frac{3}{k}})^p=1^p=1

    Hence A and B converges or diverges together.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Series convergence/divergence question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 4th 2010, 09:42 PM
  2. Replies: 1
    Last Post: May 15th 2009, 08:58 AM
  3. convergence or divergence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 22nd 2009, 07:28 AM
  4. Convergence or Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 28th 2009, 11:53 AM
  5. Replies: 8
    Last Post: April 24th 2008, 05:01 PM

Search Tags


/mathhelpforum @mathhelpforum