Results 1 to 2 of 2

Math Help - Telescoping series

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    297

    Telescoping series

    Can someone check my solution:

    Consider the series \sum_{k=1}^{\infty}a_k where a_k=<br />
\frac{1}{k(k+1)}

    1) Find the partial fractions decomposition of a_k

    a_k=\frac{1}{k}-\frac{1}{k+1}

    2) Find a simple formula for the partial sum s_n, where s_n=\sum_{k=1}^{n}a_k

    s_n=\sum_{k=1}^{n}a_k=\sum_{k=1}^{n}\frac{1}{k}-\frac{1}{k+1}

    =\sum_{k=1}^{n}\frac{1}{k}-\frac{1}{k+1}

    =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}

    =1-\frac{1}{n+1}

    3) Hence find

    \sum_{k=1}^{\infty}a_k



    \sum_{k=1}^{\infty}a_k=\lim_{n\to\infty}\sum_{k=1}  ^{n}a_k

    =\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2010
    Posts
    185
    Thanks
    13
    Looks fine from overe here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 13th 2011, 11:39 AM
  2. Telescoping series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 21st 2010, 03:18 AM
  3. Telescoping Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 16th 2010, 11:09 AM
  4. Telescoping Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 30th 2009, 09:28 AM
  5. Telescoping Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 1st 2008, 05:00 PM

Search Tags


/mathhelpforum @mathhelpforum