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Converge or Diverge?

  1. Apr 26, 2004 #1
    Does this sum converge of diverge? C is a constant

    [itex] \sum_{i=1}^{\infty} x^{C-i} [/itex]

    Is there an easy way to tell if something converges or diverges?
     
  2. jcsd
  3. Apr 26, 2004 #2

    arildno

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    Convergence of the given sum depends on the absolute value of x.

    Note that you can rewrite your sum as x^(C)*Sum(i=1,inf)1/(x^(i))
    But this is closely related to the well-known geometric series..
     
  4. Apr 26, 2004 #3
    Ok if X is a whole number, it diverges right?
     
  5. Apr 27, 2004 #4

    matt grime

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    It is a geometric progression with initial term x^c and common ratio x^{-1} (arildno has a minus sign missing}

    as such it converges for all reall x with |1/x|<1, ie |x|>1
     
  6. Apr 27, 2004 #5

    HallsofIvy

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    Since C is a constant and the sum is over the index i, you can take xC out of the sum and get
    [itex] x^C \sum_{i=1}^{\infty} x^{-i} [/itex]

    You should now be able to recognize the remaining sum as a geometric series in (1/x) which converges for -1< 1/x < 1- that is, x< -1 or x> 1.
    Convergence and divergence has nothing whatsoever to do with whether x is a whole number or not.
     
  7. Apr 27, 2004 #6
    Ok here is my next question, is it possible that:

    [itex]X^C < \sum_{i=1}^{\infty} X^{C-i} [/itex]
     
    Last edited: Apr 27, 2004
  8. Apr 27, 2004 #7

    matt grime

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    That would depend on what X, and x are. As the sum is [tex]\frac{x^{C+1}}{1-1/x}[/tex] given that for the sum to make sense |x|>1. I'm sure you can do the manipulation.

    edited to allow for the sum running from 1 to infinity, not 0 to infinity
     
    Last edited: Apr 27, 2004
  9. Apr 27, 2004 #8

    arildno

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    Hmm.., I thought the sum was x^(C)/(x-1)
     
  10. Apr 27, 2004 #9

    matt grime

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    it could well be, i get bored keeping track of the details. i now think it is x^{c-1}/(1-1/x), which was the second one i put in there and works out at x^c/(x-1) doesn't it?
     
  11. Apr 27, 2004 #10

    arildno

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    Sure, I thought it was merely a typo or someting.
     
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