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Telescoping Method & Partial Fractions PLEASE HELP

  1. Apr 19, 2008 #1
    Telescoping Method & Partial Fractions...PLEASE HELP!!!

    1. The problem statement, all variables and given/known data
    Find the sum of the series from n=1 to infinity...
    2/(4n^2-1)


    2. Relevant equations



    3. The attempt at a solution
    I want to use the telescoping method....
    2/(4n^2) = 2/[(2n-2) * (2n+1)]

    I am following an example in my book and this is where I don't know what they did next...but they got
    An = 1/(2n-1) - 1/(2n+1)
     
  2. jcsd
  3. Apr 19, 2008 #2

    Tom Mattson

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    Correct. And how do you write down the partial fraction decomposition when your denominator has two nonrepeated, linear factors?
     
  4. Apr 19, 2008 #3

    tiny-tim

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    Hi BuBbLeS01! :smile:

    I'm confused … 1/(2n - 1) - 1/(2n + 1) is 2/(4n² - 1). :confused:

    Next step … what is An + An+2 ? :smile:
     
  5. Apr 19, 2008 #4

    Tom Mattson

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    Right, he's asking how the author got that result. So he needs to revisit partial fraction decomposition.
     
  6. Apr 19, 2008 #5
    2/(4n^2) = 2/[(2n-1) * (2n+1)]
    So the LCD for the right side would be (2n-1)*(2n+1)

    2(2n+1)/(2n-1) + 2(2n-1)/(2n+1)

    For some reason that doesn't look right though??
     
  7. Apr 19, 2008 #6

    Tom Mattson

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    Hold on a second, you dropped a "-1" somewhere along the way. It should be this:

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

    That's because it's not right, and I don't understand why you're writing it. What you're supposed to do is this:

    [tex]\frac{2}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}[/tex]

    That ought to jog your memory enough to finish.
     
  8. Apr 19, 2008 #7

    tiny-tim

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    Hi BuBbLeS01!

    That's because you've put a + in the middle instead of a - . :smile:
     
  9. Apr 19, 2008 #8
    So is it just....
    2 = A(2n+1) + B(2n-1)
     
  10. Apr 19, 2008 #9
    A = 1 and B = -1
     
  11. Apr 19, 2008 #10

    tiny-tim

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    :smile: Yes! :smile:
     
  12. Apr 19, 2008 #11
    WOO-HOO LOL Thank you....I think I got it from here :)
     
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