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Telescoping Sum: 3/(n(n+3))

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series is convergent or divergent by expressing [tex]s_{n}[/tex] as a telescoping sum. If it is convergent, find its sum.
    [tex]\sum\frac{3}{n(n+3)}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Partial Fraction Decomposition: [tex]\frac{1}{n} - \frac{1}{n+3}[/tex]

    Partial Sum: [tex]s_{n}= (\frac{1}{1} - \frac{1}{4}) + (\frac{1}{2} - \frac{1}{5}) + (\frac{1}{3} - \frac{1}{6}) + (\frac{1}{4} - \frac{1}{7}) + ... + (\frac{1}{n} - \frac{1}{n+3})[/tex]

    From the above partial sum, I deduced that the negative term of the nth term is canceled out by the positive term in the n+4th term. However, from there, I am not able to come up with a general formula for the sum of the series. Any help would be appreciated; thanks!
     
  2. jcsd
  3. Sep 27, 2009 #2

    Dick

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    So which terms DON'T cancel? I count six of them for large n.
     
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