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Sum of Convergent Series Help Please!

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the sum of the convergent series:
    The sum of 1/ (n^2 - 1) from n=2 to infinity


    2. Relevant equations



    3. The attempt at a solution

    I want to break it down into 2 fractions and use partial fractions.

    1/(n-1)(n+1)....but I don't know where to go from here...
     
  2. jcsd
  3. Apr 9, 2008 #2

    HallsofIvy

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    How about doing what you just said: use partial fractions. If
    [tex]\frac{1}{n^2-1}= \frac{A}{n-1}+ \frac{B}{n+1}[/tex]
    what are A and B?
     
  4. Apr 9, 2008 #3
    A would be 1 * (n+1) = n+1
    B would be 1 * (n-1) = n-1
    right? or no??
     
  5. Apr 9, 2008 #4

    Dick

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    Noo. Take what Halls gave you and multiply both sides by (n-1)*(n+1) (remembering (n-1)*(n+1)=n^2-1). That should give you an equation to solve for A and B. What is it?
     
  6. Apr 9, 2008 #5
    So I would have
    A(n+1) * B(n-1)
     
  7. Apr 9, 2008 #6

    Dick

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    No. That's not even an equation. It should 'equal' something. And how did you get the '*' between the two terms. Try again. Show more work if you are really confused.
     
  8. Apr 9, 2008 #7
    Ok that should have been + not * my mistake...
    1/n^2 - 1 = A/n-1 + B/n+1
    1/n^2 - 1 = A(n-1)(n+1)/n-1 + B(n-1)(n+1)/n+1
    1/n^2 - 1 = A(n+1) + B(n-1)
     
  9. Apr 9, 2008 #8

    Dick

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    You are getting closer, but you didn't multiply the left side by (n^2-1), did you?
     
  10. Apr 9, 2008 #9
    Oh I didn't realize I had to....
    So I get...
    1 = A(n+1) + B(n-1)
     
  11. Apr 9, 2008 #10

    Dick

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    You've got it. Now write that as 1=(A+B)*n+(A-B). There's no 'n' on the left side so the coefficient of n on the right side should be zero, and the constant term should be 1. Can you find A and B?
     
  12. Apr 9, 2008 #11
    Can I also just do...
    1 = A(n+1) + B(n-1)
    Plug in 1 to solve for A and -1 to solve for B? Getting 1/2 and -1/2???
     
  13. Apr 9, 2008 #12

    Dick

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    Sure. Plugging in values of n will work too. But plugging in just n=1 doesn't tell you what B is, does it?
     
  14. Apr 9, 2008 #13
    can't he plug in -1 for to zero out A and get B

    (heaviside cover up method?)
     
  15. Apr 9, 2008 #14
    Yea thats what I meant 1 to get A and -1 to get B
     
  16. Apr 9, 2008 #15

    Dick

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    I said, sure you can! You could also, without paying much attention to n, just say (A+B)=0, and (A-B)=1 and solve them simultaneously. Whichever method you prefer. They both work. Now can you sum the series?
     
  17. Apr 10, 2008 #16
    1/n^2 - 1 = .5/n-1 - .5/n+1

    Then I start with plugging in 2 going to infinity...

    [.5/2-1 - .5/2+1] + [.5/3-1 - .5/3+1] + [.5/4-1 - .5/4+1] + [.5/5-1 - .5/5+1].......and so on...

    [.5/1 - .5/3] + [.5/2 - .5/4] + [.5/3 - .5/5] + [.5/4 - .5/6].......

    Some cancel out...I thought it was supposed to be all of them cancel except the 1st and nth term? Somehow I didn't get that...
     
  18. Apr 10, 2008 #17

    Dick

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    The next term that you didn't write will cancel the .5/5. The one after that will cancel the .5/6. The one after that will cancel the .5/7. Etc. In the end, which terms don't cancel?
     
  19. Apr 10, 2008 #18
    I thought it was supposed to be the first and last but then .5/2 didn't cancel which is half of the 2nd term???
     
  20. Apr 10, 2008 #19

    Dick

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    I'm not sure why you are saying 'first' and 'last' should cancel, but you are right. .5/2 doesn't cancel.
     
  21. Apr 10, 2008 #20
    Strictly speaking a telescoping series doesn't have to be just the first and last term, as long as enough terms cancel then that's all that matters.
     
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