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Showing the sum of this telescoping series

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Determine whether each of the following series is convergent or divergent. If the series is convergent, find its sum

    [tex] \sum_{i=1}^{\infty} \frac{6}{9i^{2}+6i-8} [/tex]

    2. Relevant equations

    Partial fraction decomposition

    [tex] \frac{1}{3i-2} - \frac{1}{3i+4} [/tex]

    3. The attempt at a solution

    The divergence test is inconclusive, so I wrote as partial fractions and started analysing the nth sum:

    [tex] S_{n} = \left( 1-\frac{1}{7} \right) + \left( \frac{1}{4} - \frac{1}{10} \right) + \left( \frac{1}{7} - \frac{1}{13} \right) + \left( \frac{1}{10} - \frac{1}{16} \right) + \dots [/tex]

    1 and 1/4 are the only terms that do not cancel, but how do I show this in the nth case? I'm having trouble writing it generally.
     
  2. jcsd
  3. Feb 2, 2016 #2

    Mark44

    Staff: Mentor

    Include the general term in your expansion:
    ##S_{n} = \left( 1-\frac{1}{7} \right) + \left( \frac{1}{4} - \frac{1}{10} \right) + \left( \frac{1}{7} - \frac{1}{13} \right) + \left( \frac{1}{10} - \frac{1}{16} \right) + \dots + \left( \frac{1}{3n-2} - \frac{1}{3n+4} \right) + \dots##
    If you add in the term before and the one after the last term I wrote above, you should see how the telescoping happens.
     
  4. Feb 2, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Show that for every term of the form ##1/(2n)## there will be another term ##-1/(2n)## (corresponding to just two possible values of ##i##) and for every term ##1/(2n+1)## there is a cancelling term ##-1/(2n+1)##---again, corresponding to exactly two values of ##i##.
     
  5. Feb 2, 2016 #4
    Thanks guys, I got it now.
     
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