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Geometric Series

  1. Mar 10, 2010 #1
    {sigma} 2*(-1/4)^(n-1)

    Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

    my A=2
    my r= -1/4
     
  2. jcsd
  3. Mar 10, 2010 #2

    Mark44

    Staff: Mentor

    Your series appears to be
    [tex]\sum_{n = 1}^{\infty} 2 (-1/4)^{n - 1}[/tex]

    Write a few terms of this series and see if you can adjust the starting index.
     
  4. Mar 10, 2010 #3
    i have no idea what you mean by that
     
  5. Mar 10, 2010 #4

    Hurkyl

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    Gold Member

    You have
    2(-1/4)n-1
    and you want
    A rn-1
    ?

    Why not set them equal and try solving for A and r?
     
  6. Mar 10, 2010 #5
    no... im trying to solve whether or not the series converges or diverges and then i have to determine the sum of the series... meaning where the series sums up to..
     
  7. Mar 10, 2010 #6

    Hurkyl

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    Oh, my mistake. I thought the method you had wanted to use was to rewrite the summand in a standard form, and were stuck on how to do that rewrite.
     
  8. Mar 10, 2010 #7

    Char. Limit

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    Gold Member

    Well, just change n-1 into n using exponent rules.
     
  9. Mar 10, 2010 #8

    Mark44

    Staff: Mentor

    Write a few terms of the series and you should quickly see that it is a geometric series. A geometric series is one for which each term is a constant multiple of the previous term.
     
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