Results 1 to 2 of 2

Thread: The sum of another problematic sequence.

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    The sum of another problematic sequence.



    That's the question and my working so far. I can't really see what the next step is or if this method will work out at the end.

    Can someone please help me?
    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
    9
    Quote Originally Posted by Showcase_22 View Post


    That's the question and my working so far. I can't really see what the next step is or if this method will work out at the end.

    Can someone please help me?
    Go back to the original series and substitute $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ ....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. problematic integral #2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 21st 2010, 02:54 AM
  2. problematic integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 20th 2010, 08:31 AM
  3. problematic limit
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Jan 10th 2010, 11:43 AM
  4. Problematic Inverses
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 11th 2009, 07:20 AM
  5. The sum of a most problematic sequence.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Sep 20th 2008, 10:01 AM

Search Tags


/mathhelpforum @mathhelpforum