Results 1 to 3 of 3
Like Tree2Thanks
  • 1 Post By SlipEternal
  • 1 Post By Prove It

Math Help - Find the sum of a series

  1. #1
    Newbie
    Joined
    Aug 2013
    From
    Quebec
    Posts
    5

    Find the sum of a series

    Hello there,

    I'm a litttle bit stuck on a math question here

    I need to find the Sum of the serie 1/((4*i^2)-1)
    Find the sum of a series-sum-problem.png *made a mistake in the picture the i element is suppose to be a n sorry

    I can't seem to understand how to even start

    thank you guys
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,930
    Thanks
    782

    Re: Find the sum of a series

    Let's look at partial sums: S_k = \sum_{n=1}^k \dfrac{1}{4n^2-1}. We have

    S_1 = \dfrac{1}{3}

    S_2 = \dfrac{1}{3} + \dfrac{1}{15} = \dfrac{2}{5}

    S_3 = \dfrac{2}{5} + \dfrac{1}{35} = \dfrac{3}{7}

    In general, the pattern appears to be S_k = \dfrac{k}{2k+1}. You can show inductively that this is the case.

    So, \sum_{n=1}^\infty \dfrac{1}{4n^2-1} = \lim_{k \to \infty} \dfrac{k}{2k+1} = \dfrac{1}{2}.
    Thanks from Memorystack
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,654
    Thanks
    1478

    Re: Find the sum of a series

    Quote Originally Posted by Memorystack View Post
    Hello there,

    I'm a litttle bit stuck on a math question here

    I need to find the Sum of the serie 1/((4*i^2)-1)
    Click image for larger version. 

Name:	Sum problem.png 
Views:	11 
Size:	1.1 KB 
ID:	29527 *made a mistake in the picture the i element is suppose to be a n sorry

    I can't seem to understand how to even start

    thank you guys
    Notice that \displaystyle \begin{align*} \frac{1}{4n^2 - 1} = \frac{1}{(2n - 1)(2n + 1)} \end{align*}

    Split it up into partial fractions and you should find that this is a telescopic series.
    Thanks from SlipEternal
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: December 9th 2012, 07:12 PM
  2. Replies: 2
    Last Post: December 9th 2012, 12:50 PM
  3. Replies: 2
    Last Post: May 22nd 2012, 05:57 AM
  4. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  5. Binomial Series to find a Maclaurin Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 21st 2009, 07:15 AM

Search Tags


/mathhelpforum @mathhelpforum