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Finding the sum of the series

  1. May 9, 2012 #1
    So the question is find the sum of the series:


    [itex]\sum[/itex] from k = 1 to ∞ of 1 / k(k+3),

    now the solution they provided was:

    = 1/3 [ (1 - 1/4) + (1/2 - 1/5) + ...+ (1/n - 1/(n+3) ]

    =1/3 [ 1+ 1/2+1/3 -1/(n+1) - 1/(n+2) - 1/(n+3)]

    --> 11/18

    I'm stuck on how they were able to show the sum telescoping and why they were able to factor out the 1/3. Also how are you suppose to solve these sorts of questions?
     
  2. jcsd
  3. May 9, 2012 #2

    Mark44

    Staff: Mentor

    Rewrite 1/(k(k + 3)) as two fractions, using partial fraction decomposition.
     
  4. May 9, 2012 #3

    sharks

    User Avatar
    Gold Member

    [tex]\sum^{\infty}_{k=1} \frac{1}{k(k+3)}[/tex]
    First, express [itex]\frac{1}{k(k+3)}[/itex] as partial fractions.
     
  5. May 9, 2012 #4
    Darn. Have to go back and review that, but it makes sense to do that now, thanks.
     
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