Results 1 to 4 of 4

Math Help - Stuck on evaluating a series

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    21

    Stuck on evaluating a series

    How would evaluate the following series?

    \sum _{i=3}^{\infty }2\,{\frac {1}{ \left( i-2 \right)  \left( i+2<br />
 \right) }}

    I can get my calculator to evaluate this, but I am stuck when trying to calculate it manually. Is this a telescoping series where I would need to use partial fractions? I can't get it to work out that way.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by RU2010 View Post
    How would evaluate the following series?

    \sum _{i=3}^{\infty }2\,{\frac {1}{ \left( i-2 \right)  \left( i+2<br />
 \right) }}

    I can get my calculator to evaluate this, but I am stuck when trying to calculate it manually. Is this a telescoping series where I would need to use partial fractions? I can't get it to work out that way.
    It is not quite a telescoping series, but do use partial fractions and write out the first few terms and you will see what actually cancels out.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,507
    Thanks
    1403
    These are nearly always able to be evaluated as a telescopic series after a partial fraction decomposition.

    \frac{A}{i - 2} + \frac{B}{i + 2} = \frac{1}{(i - 2)(i + 2)}

    \frac{A(i + 2) + B(i - 2)}{(i - 2)(i + 2)} = \frac{1}{(i - 2)(i + 2)}

    A(i + 2) + B(i - 2) = 1

    Ai + 2A + Bi - 2B = 1

    (A + B)i + 2A - 2B = 0i + 1.


    A+B=0 and 2A - 2B = 1.

    Substituting B = -A into the second equation gives

    2A + 2A = 1

    4A = 1

    A = \frac{1}{4}.

    Therefore B= -\frac{1}{4}.


    So finally we have

    \frac{1}{(i - 2)(i + 2)} = \frac{1}{4(i - 2)} - \frac{1}{4(i + 2)}.


    Therefore our series is

    2\sum_{i = 3}^{\infty}\frac{1}{(i - 2)(i + 2)} = 2\lim_{n \to \infty}\sum_{i = 3}^{n}\left[\frac{1}{4(i - 2)} - \frac{1}{4(i + 2)}\right]

     = \frac{1}{2}\lim_{n \to \infty}\sum_{i = 3}^{n}\left(\frac{1}{i - 2} - \frac{1}{i + 2}\right)

     = \frac{1}{2}\lim_{n \to \infty}\left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{2} - \frac{1}{6}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \left(\frac{1}{5} - \frac{1}{9}\right) + \left(\frac{1}{6} - \frac{1}{10}\right) + \left(\frac{1}{7} - \frac{1}{11}\right) + \dots + \left(\frac{1}{n - 2} - \frac{1}{n + 2}\right)

     = \frac{1}{2}\lim_{n \to \infty}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{n - 1} - \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{n + 2}\right) since everything cancels except the first and last four terms

     = \frac{1}{2}\lim_{n \to \infty}\left(\frac{25}{12} - \frac{1}{n - 1} - \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{n + 2}\right)

     = \frac{1}{2}\left(\frac{25}{12}\right)

     = \frac{25}{24}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2010
    Posts
    21
    Got it! Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Getting stuck on series - can't develop Taylor series.
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 5th 2010, 08:32 AM
  2. [SOLVED] I'm stuck evaluating this limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 18th 2010, 12:38 PM
  3. Evaluating series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 10th 2010, 06:06 PM
  4. Evaluating a Series of Binomial Co-efficients
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 19th 2009, 04:46 AM
  5. Evaluating Series.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 14th 2008, 05:30 PM

Search Tags


/mathhelpforum @mathhelpforum