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Geometric Series

  1. Jul 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Does the series from n=1 to infinity of (2)/(n^2-1) converge or diverge? If it converges, find the sum.

    2. Relevant equations



    3. The attempt at a solution

    I can see right away that the series converges by a limit comparison test by looking at the series. However, to find the sum I have re-write that as a geometric series. There is nothing, at least to me, that gives away how to re-write that as a geometric series. That's where I'm stuck.

    Thanks for any help.
     
  2. jcsd
  3. Jul 16, 2008 #2

    rock.freak667

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    Why don'y you write [itex]\frac{2}{n^2-1}[/itex] in partial fractions and see if it is a telescoping series?


    Find

    [tex]\sum_{n=1} ^{N} \frac{2}{n^2-1}[/tex]

    and then check what happens as [itex]N \rightarrow \infty[/itex]
     
  4. Jul 16, 2008 #3
    I see it now. Thank you.
     
  5. Aug 25, 2008 #4
    what's the summation of the series?

    what's the summation of the series?
    1.A^1 + 2.A^2 + 3.A^3 + .............+n.A^n.

    please reply
     
  6. Aug 25, 2008 #5

    rock.freak667

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    Re: what's the summation of the series?


    What have you tried so far on it?
     
  7. Aug 25, 2008 #6

    Defennder

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    Re: what's the summation of the series?

    Is this related to the original post or something separate?
     
  8. Aug 26, 2008 #7

    statdad

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    Are you sure you don't mean

    [tex]
    \sum_{n=2}^\infty \frac{2}{n^2-1}
    [/tex]

    i.e.- the sum starting at [tex] n = 2 [/tex]? If you try to start at [tex] n = 1 [/tex] the very first term is undefined (can't divide by zero) so the series would not converge. As I mentioned - this makes the difference between the series converging and not converging, and will influence your value for the sum.
     
  9. Aug 26, 2008 #8

    Gib Z

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    Yes, he probably meant the sum to start at n=2.

    Jainal36- you should really start new threads rather than hijacking others, but ill help anyway - thats just A times the derivative of a series you know how to sum.
     
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