Results 1 to 3 of 3

Math Help - just wanna add another one

  1. #1
    Senior Member
    Joined
    Jan 2010
    Posts
    273

    just wanna add another one

    \sum\limits_{k = 1}^\infty 1/[2k(k+1)]


    \sum\limits_{k = 1}^\infty 3^{k-1}/4^{3k+1}


    find the sum of the series

    I got the last one is 61/ (12*64), but doubt the answer, just want to confirm it
    Last edited by wopashui; March 29th 2010 at 08:26 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by wopashui View Post
    \sum\limits_{k = 1}^\infty 1/[2k(k+1)]

    find the sum of the series
    It's a telescoping series. Express \frac{1}{k (k+1)} in partial fraction form.

    (You should have learned something from your thread here and compared it to the first question you asked: http://www.mathhelpforum.com/math-he...um-series.html)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,977
    Thanks
    1121
    Quote Originally Posted by wopashui View Post
    \sum\limits_{k = 1}^\infty 1/[2k(k+1)]


    \sum\limits_{k = 1}^\infty 3^{k-1}/4^{3k+1}


    find the sum of the series

    I got the last one is 61/ (12*64), but doubt the answer, just want to confirm it
    \frac{3^{k-1}}{4^{3k+1}}= \frac{3^k(3^{-1}}{4^{3k}(4)} = \frac{1}{12}\frac{3^k}{(4^3)^k}= \frac{1}{12}\left(\frac{3}{64}\right)^k so the second is a geometric series with first term \frac{1}{12} and common ratio \frac{3}{64}.

    I think you have a fraction upside down.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I just wanna say thanks..
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 1st 2009, 08:01 PM
  2. Replies: 8
    Last Post: August 24th 2007, 11:46 AM
  3. someone wanna look over my work?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 11th 2006, 08:54 PM

Search Tags


/mathhelpforum @mathhelpforum