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Partial Fractions Sum of Series

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Use partial fractions to find the sum of the series.


    2. Relevant equations
    [itex]\displaystyle \sum^{∞}_{n=1} \frac{8}{n(n+3)} [/itex]


    3. The attempt at a solution
    I end up with:

    [itex]\displaystyle \frac{8}{3n} - \frac{8}{3(n+3)}[/itex]

    I am stuck here.
     
  2. jcsd
  3. Apr 4, 2013 #2
    Did you try writing out some terms of the series? I took it to n=7 to get a good feeling.
     
  4. Apr 4, 2013 #3

    tiny-tim

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    hi whatlifeforme! :smile:

    ok, now try writing out the first few terms, and see what you notice :wink:

    (if that doesn't help, try it with n(n+1) instead of n(n+3), and then adapt)
     
  5. Apr 4, 2013 #4
    i'm still confused. after 7 terms, i get to 3.99 but it looks like it is still increasing.
     
  6. Apr 4, 2013 #5
    Ack, decimals are bad :) Use actual fractions, and write it out like this:
    [tex] \frac{8}{3}-\frac{8}{12}+\frac{8}{6}-\frac{8}{15}+\dots [/tex]
    Continue this pattern to at least n=7. You might want to look up "telescoping series".
     
  7. Apr 4, 2013 #6
    i have yet to master telescoping series. any help please?

    I have gone to n=7 and have 6 terms left that have not be canceled.
     
  8. Apr 4, 2013 #7

    tiny-tim

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    and what are the 6 terms? :smile:
     
  9. Apr 4, 2013 #8
    8/3 + 8/6 + 8/9 - 8/24 - 8/27 - 8/30
     
  10. Apr 4, 2013 #9

    Ray Vickson

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    Evaluate the finite sum
    [tex]S_N = \sum_{n=1}^N \frac{8}{n(n+3)},[/tex] then take the limit as N → ∞. Your partial fraction representation makes this straightforward.
     
  11. Apr 4, 2013 #10

    SammyS

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    Write them out without cancelling them.
     
  12. Apr 4, 2013 #11
    (8/3 - 8/13) + (8/6 - 8/15) + (8/9 -8/15) + (8/12 - 8/21) + (8/15 - 8/24) + (8/18 - 8/27) + (8/21 - 8/30)
     
  13. Apr 4, 2013 #12
    Here is an example. Suppose my series came out like this:
    [tex] \frac{1}{2} - \frac{1}{5} +\frac{1}{3} - \frac{1}{6} + \frac{1}{4} - \frac{1}{7} + \frac{1}{5} - \frac{1}{8} + \frac{1}{6} + \dots [/tex]

    Notice that this can be rearranged as [tex] \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{5} - \frac{1}{5} +\frac{1}{6} - \frac{1}{6} \dots [/tex]
    See that every fraction has a positive and a negative, so everything cancels except for the first few fractions, 1/2, 1/3, and 1/4. The sum of this series is [tex] \frac{1}{2}+\frac{1}{3}+\frac{1}{4} [/tex].
     
  14. Apr 4, 2013 #13
    Can you form pairs of positive and negative like -8/15 + 8/15 ?
     
  15. Apr 4, 2013 #14

    SammyS

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    How about like this ?

    8/3 + 8/6 + 8/9 + 8/12 + 8/15 + 8/18 + 8/21 + ...

    - 8/12 - 8/15 - 8/18 - 8/21 - 8/24 - 8/27 - 8/30 - ...

    Do you see what happens?
     
  16. Apr 4, 2013 #15
    so i'm left with 8/3 + 8/6 + 8/9
     
  17. Apr 4, 2013 #16
    Exactly :) Does it make sense?
     
  18. Apr 4, 2013 #17
    not yet. i still don't have the sum. doesn't it need to look like sum = 8/3 - 1/k-1

    then taking the limit it would be 8/3.

    this isn't valid for this problem just an example.

    i know when writing out 1/n - 1/n-1 type stuff.. i always had a problem with telescoping series.
     
  19. Apr 4, 2013 #18
    The sum is a number. Specifically the sum is the sum of 8/3+8/6+8/9.
     
  20. Apr 4, 2013 #19
    so the answer is 44/9.
     
  21. Apr 4, 2013 #20
    Right.
     
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