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Math Help - Eventually self-outgrowing = eventually monotonic?

  1. #1
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    Eventually self-outgrowing = eventually monotonic?

    Hi,

    Consider a bounded sequence (a_n)_{n \in \mathbb{N}} whose terms are eventually "outgrown" by other terms of the sequence, i.e., for any index n there exists an n' > n such that a_{n'} > a_n. Does this property imply that a_n is eventually monotonic, i.e., that there exists an index n_0 such that a_n is monotonically increasing from index n_0 onwards? I can't think of any counterexample. If true, do you know if this a known theorem which I can reference from some standard textbook?

    NB: the boundedness assumption is essential, otherwise you can construct simple counterexamples like a_n = n + (-1)^n

    Thanks,
    jens
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  2. #2
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    Re: Eventually self-outgrowing = eventually monotonic?

    a_{2n}:=1-\frac 1{2n} and a_{2n+1}=0.
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  3. #3
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    Re: Eventually self-outgrowing = eventually monotonic?

    Thanks for your counterexample!
    Sorry for the mess, but I had only sequences in mind whose variations become "tiny", that is, whose \lim\sup and \lim\inf coincide. But I forgot to tell. So let me correct my question:
    Assume a_n converges to a finite limit and for any index n, there exists n' > n such that a_{n'} > a_n. Does this imply that there exists an index n_0 such that for any two n_1 and n_2 both larger than n_0, we have n_1 < n_2 \Rightarrow a_{n_1} < a_{n_2}?
    Now I hope this statement is correct (and non-trivial!)
    Last edited by jens; May 31st 2012 at 02:52 PM.
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  4. #4
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    Re: Eventually self-outgrowing = eventually monotonic?

    What about a_{2n}=1-1/n^2 and a_{2n+1}=1-1/n?
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  5. #5
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    Re: Eventually self-outgrowing = eventually monotonic?

    Ok, thanks!! So my conjecture was hopelessly wrong…
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