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Sequence proof

  1. Oct 3, 2005 #1
    Hi,
    Here is the question:
    Prove that if the sequence {s} has no convergent subsequence then {|s|} diverges to infinity.

    To me, this seems so easy, but I'm having a really hard time putting it down in a rigorous manner.
    My thoughts are:
    every convergent sequence has a convergent subsequence (theorem in the book), so if there is no convergent subsequence, then the sequence itsself cannot converge either.
    So the absolute value won't converge.

    I tried to do it with contradiction as follows:
    Suppose that |s| converges. Then it has a convergent subsequence. Since |s| is also a subsequence of {s}, the convergent subsequence of |s| lies inside {s}. So {s} has a convergent subsequence. Contradiction.

    Any clarification will be greatly appreciated.
    CC
     
  2. jcsd
  3. Oct 4, 2005 #2

    CarlB

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    There is a problem with your logic in that {|s|} is not necessarily a subsequence of {s}. This is true even for convergent series like:

    [tex] s_n = \frac{(-1)^n}{n}[/tex]

    Carl
     
  4. Oct 4, 2005 #3
    Ok,
    I see that. Can I say that the elements a convergent subsequence of |s| also are members of {s}? The convergent subsequence of |s| will be positive numbers converging to L, so {s} will either have the same exact subsequence of positive terms, or it'll be negative terms, but then will converge to -L
     
  5. Oct 4, 2005 #4

    CarlB

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    happyg1,

    Was the original problem "Prove that if the sequence {s} has no convergent subsequence then {|s|} diverges to infinity.", or was it instead "Prove that if the sequence {s} has no convergent subsequence then [itex] \sum |s|[/itex] diverges to infinity."

    The reason I ask this is because I'm not really sure what "diverges to infinity" means.

    If the meaning is to say that given a number L, no matter how far down the series you look, you can find an element larger in absolute value than L, then I can see your terminology working.

    The whole thing smells of recourse to an argument based on the fact that any sequence that stays inside a closed subset of the real line has a convergent subsequence.

    Carl
     
  6. Oct 4, 2005 #5
    Prove that if the sequence {s} has no convergent subsequence then {|s|} diverges to infinity.

    that is the question as printed in my book.
     
  7. Oct 4, 2005 #6

    CarlB

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    Then I think you will need to make an argument based on a fact that I vaguely recall to the effect that any sequence to a compact space has a subsequence that converges to an element of the compact space.

    Hey, it's been about 30 years since I studied this, and this is the limit of what I can help.

    Carl
     
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