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Sum of a series question

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the series.

    [tex] \sum_{k=0}^\infty \frac{1}{(k+1)(k+3)} [/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex]
    = \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \frac{1}{3\cdot5} + ... + \frac{1}{(n+1)\cdot(n+3)}
    [/tex]

    [tex]
    = \frac{1}{2} [(1-\frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + ... + (\frac{1}{(n+1)} - \frac{1}{(n+3)})
    [/tex]

    [tex]
    = \frac{1}{2}[ 1 + (\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) - (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) ]
    [/tex]

    So here
    [tex]
    (\frac{1}{2} + \frac{1}{3} + \frac{1}{n+1}) \to 1
    [/tex]

    [tex]
    (\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... + \frac{1}{n+3}) \to 0
    [/tex]

    Then the whole thing sums to 1?
     
  2. jcsd
  3. Mar 23, 2010 #2
    Ok, things went very bad from the second to the third line in your argument; why don't you try to cancel the terms with opposite signs, instead of grouping them?
     
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