Need help with upper limit of sequence.

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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^*
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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