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Finding the formula for the partial sum Sn

  1. Jul 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the series Ʃ 1/[k(k+2)]; n=1 to infinity
    Find the formula for the partial sum Sn


    2. The attempt at a solution
    I have calculated the first 5 terms of the sequence as follows, but I can't see any pattern. Am I doing this right?
    S1=1/3
    S2=1/3+1/8=11/24
    S3=1/3+1/8+1/15=21/40
    S4=1/3+1/8+1/15+1/24=17/30
    S5=1/3+1/8+1/15+1/24+1/35=25/42
    Sn=??? I can't find any ratio or exponential relation between the terms. I need to find a general formula for Sn
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 27, 2013 #2
    Hint: [itex] \frac{1}{k(k+2)}=\frac{1}{k}-\frac{1}{k+2} [/itex]. Your series is telescoping.
     
  4. Jul 27, 2013 #3
    Telescoping series

    Ok I have read a bit on telescoping series. One thing I do not understand is this:

    Why does it go from 1/[k(k+2)] to 1/k - 1/k+2, why are they subtracted instead of added?
     
  5. Jul 27, 2013 #4

    LCKurtz

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    What happens if you simplify ##\frac 1 k - \frac 1 {k+2}##? Does it equal ##\frac 1 {k(k+2)}## or not? That will answer your question.
     
  6. Jul 28, 2013 #5

    ehild

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    It is [itex] \frac{1}{k(k+2)}=0.5\left(\frac{1}{k}-\frac{1}{k+2}\right)[/itex]

    ehild
     
  7. Jul 28, 2013 #6

    Curious3141

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    Have you learnt about partial fractions?
     
  8. Jul 28, 2013 #7

    HallsofIvy

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    More to the point, don't you know how to add and subtract fractions?
    To add or subtract 1/k and 1/(k+2) (the parentheses in that second fraction are important!) you need to get the same denominator which is, of course, k(k+2).

    1/k+ 1/(k+2)= (k+2)/k(k+2)+ k/k(k+2)= (2k+2)/k(k+2)

    1/k- 1/(k+2)= (k+2)/k(k+2)- k/k(k+2)= 2/k(k+2)

    Which is the one you want?
     
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