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I Geometric Series Confusion

  1. Apr 19, 2016 #1

    Drakkith

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    Staff: Mentor

    I'm a little confused on geometric series.

    My book says that a geometric series is a series of the type: n=1 to ∞, ∑arn-1
    If r<1 the series converges to a/(1-r), otherwise the series diverges.

    So let's say we have a series: n=1 to ∞, ∑An, with An = 1/2n
    An can be re-written as (1/2)n, which apparently makes it a geometric series with r=1/2. This converges to 1/(1-1/2) = 1/(1/2) = 2.

    However, I was under the assumption that I was supposed to factor out a 1/2 to make the series ∑1/2(1/2)n-1.
    This would converge to (1/2)/(1-1/2) = (1/2)/(1/2) = 1.

    Why are these different? Which one is correct?
     
  2. jcsd
  3. Apr 19, 2016 #2

    Samy_A

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    ##\displaystyle \sum_{n=1}^\infty \frac{1}{2^n} = \frac{1}{2} +\frac{1}{4}+\frac{1}{8}+ \cdots =1##
    ##\displaystyle \sum_{n=1}^\infty \frac{1}{2^{n-1}} =1 +\frac{1}{2} +\frac{1}{4}+\frac{1}{8}+ \cdots=2##
     
  4. Apr 19, 2016 #3

    Math_QED

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    The formule for the total infinite sum is: u1/(1-r)

    So, for the first sum, u1 = 1/2, and r = 1/2, so (1/2)/(1-1/2) = 1 and not 2.

    So both series are equal.
     
  5. Apr 22, 2016 #4

    Mark44

    Staff: Mentor

    We have some very nice MathJax notation for this :oldbiggrin:
    $$\sum_{n = 1}^{\infty}ar^{n - 1}$$

    The script I wrote is \sum_{n = 1}^{\infty}ar^{n - 1}, with a pair of $ characters at the beginning and end.
     
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