Need help with upper limit of sequence.

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Homework Help Overview

The discussion revolves around the upper limit of a sequence, specifically focusing on the definition and implications of the limit superior, denoted as s^{*}. The original poster seeks to prove that s^{*} is the limit of the supremum of the sequence as n approaches infinity, under the assumption that s^{*} is finite.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to clarify the meaning of \sup_{k \geq n} s_k and its implications for the boundedness of the sequence. They express uncertainty about the application of Theorem 3.17 and how it relates to their proof. Other participants provide definitions and confirm the understanding of the supremum in this context.

Discussion Status

The discussion is active, with participants exploring the definition of the supremum and its relevance to the problem. Some guidance has been offered regarding the interpretation of the supremum, and there is a developing understanding of how the limit as n approaches infinity relates to the upper limit of the sequence.

Contextual Notes

The original poster assumes that s^{*} is finite, which leads to the conclusion that the sequence is bounded above. There is also a reference to a theorem that may be relevant, but its application remains uncertain among participants.

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Homework Statement


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

Homework Equations


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


The Attempt at a Solution


I started out writing what I know.
By assuming s^{*} is finite, then \{s_k\} is bounded above so a supremum exists.
I'm unclear what exactly \sup_{k \geq n} s_k 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?
 
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Any hints?
 
\displaystyle\sup_{k ≥ n} s_k 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.
 
Zondrina said:
\displaystyle\sup_{k ≥ n} s_k 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 E = \{s_k\} ^{\infty}_{k=n}. Since s^* is finite, then E is bounded from above, E \subset \{s_k\} and E is not empty, then a supremum exists in E. And then taking limit as n \rightarrow \infty, E would consist of only supE which is exactly s^*
 

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