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Analysis Proof

  1. Sep 16, 2007 #1
    Hello, got a proof question for you.

    QUESTION

    Prove that if an infinite series converges, then the associative property holds.



    Am I missing something here because I don't see much to this proof. In short, if the convergent series is summed in any order and does not converge to the same limit as the series would when summed in order, then the terms in the series must be different since we know the associative property holds for addition.

    What am I proving?
     
  2. jcsd
  3. Sep 16, 2007 #2

    Gib Z

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    Basically theres a property that for some divergent series, some arrangements of the terms will make it convergent, falsely. However, for all absolutely convergent series, it doesn't matter how you arrange the terms, the sum is still always the same. I believe this is what they want you to show.
     
  4. Sep 16, 2007 #3

    CompuChip

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    For example, the series
    [tex]\sum_{n = 1}^{\infty} \frac{(-1)^n}{n^2}[/tex]
    converges absolutely, as
    [tex]\sum_{n = 1}^{\infty} \left| \frac{(-1)^n}{n^2} \right| = \frac{\pi^2}{6}[/tex].

    On the other hand, the series
    [tex]\sum_{n = 0}^{\infty} (-1)^n = 1 - 1 + 1 - 1 + \cdots[/tex]
    does not converge absolutely. Indeed, summing it as
    [tex]1 - 1 + 1 - 1 + \cdots = (1 - 1) + (1 - 1) + \cdots = 0 + 0 + \cdots = 0[/tex]
    gives something different from
    [tex]1 - 1 + 1 - 1 + \cdots = 1 + (-1 + 1) + (-1 + 1) + \cdots = 1 + 0 + 0 + \cdots = 1[/tex].
    In fact, one can make it "converge" to any number one likes.
     
  5. Sep 16, 2007 #4
    Thank you. Reading the next question in my book, I see it leading to the point both of you made regarding divergent series.
     
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