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Need help with upper limit of sequence.

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that,
    [tex]s^{*} = \lim_{n \rightarrow \infty} \sup_{k \geq n} s_k[/tex]
    Assume that [itex]s^{*}[/itex] is finite.

    2. Relevant equations
    Definition of [itex]s^{*}[/itex] is here: http://i.imgur.com/AWfOW.png


    3. The attempt at a solution
    I started out writing what I know.
    By assuming [itex]s^{*}[/itex] is finite, then [itex]\{s_k\}[/itex] is bounded above so a supremum exists.
    I'm unclear what exactly [tex]\sup_{k \geq n} s_k[/tex] means. Fixing n and finding supremum of {s_k} for k >= n and then letting n -> oo? I would think if there is an upper limit for {s_k} and for all n < k, as n ->oo then {s_n} will converge to that upper limit. And I have a feeling the Theorem 3.17 in the image might be applicable to this problem?
     
  2. jcsd
  3. Nov 10, 2012 #2
    Any hints?
     
  4. Nov 10, 2012 #3

    Zondrina

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    Homework Helper

    [itex]\displaystyle\sup_{k ≥ n} s_k[/itex] means that you want to take the supremum of the set sk generated by all numbers which are greater than or equal to your n.
     
  5. Nov 10, 2012 #4
    Ok that makes sense. Then from there I could say:

    Let [itex]E = \{s_k\} ^{\infty}_{k=n}[/itex]. Since [itex]s^*[/itex] is finite, then [itex]E[/itex] is bounded from above, [itex]E \subset \{s_k\}[/itex] and [itex]E[/itex] is not empty, then a supremum exists in [itex]E[/itex]. And then taking limit as [itex]n \rightarrow \infty[/itex], [itex]E[/itex] would consist of only sup[itex]E[/itex] which is exactly [itex]s^*[/itex]
     
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