Need help with upper limit of sequence.

  • Thread starter c0dy
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  • #1
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Homework Statement


Prove that,
[tex]s^{*} = \lim_{n \rightarrow \infty} \sup_{k \geq n} s_k[/tex]
Assume that [itex]s^{*}[/itex] is finite.

Homework Equations


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


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?
 

Answers and Replies

  • #2
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Any hints?
 
  • #3
STEMucator
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[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.
 
  • #4
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[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.
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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