The Limit Superior and Bounded Sequences

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Homework Statement



\{x_{n}\}\in\mathbb{R^{+}} is a bounded sequence and r=\lim\sup_{n\rightarrow\infty}x_{n}. Show that \forall\epsilon>0,\exists finitely many x_{n}>r+\epsilon and infinitely many x_{n}<r+\epsilon.

The Attempt at a Solution



By definition of limit superior, r\in\mathbb{R} is such that \forall\epsilon>0, \exists N_{\epsilon} s.t. x_{n}<r+\epsilon, \forall n>N_{\epsilon}. This would imply that any x>r+\epsilon/is an upper bound on \{x_{n}\}. How do I show that there are finitely many such upper bounds? Is it because \{x_{n}\} is a bounded sequence that there must only be finite x_{n}>r+\epsilon ?
 
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How do I show that there are finitely many such upper bound

I'm not sure what you mean here, but what you've written basically answers the first part of the question. If for n>N_{\epsilon}, x_n<r+\epsilon, then there are at most N_{\epsilon} values of xn which are larger than r+\epsilon
 
Office_Shredder said:
I'm not sure what you mean here, but what you've written basically answers the first part of the question. If for n>N_{\epsilon}, x_n<r+\epsilon, then there are at most N_{\epsilon} values of xn which are larger than r+\epsilon

I meant that "there are finitely many such x\in(x_{n})". How do I get started proving that there are infinitely many x\in(x_{n}) s.t. x<r+\epsilon?
 
autre said:
How do I get started proving that there are infinitely many x\in(x_{n}) s.t. x<r+\epsilon?

Suppose there were only finitely many x_n such that x_n < r + \epsilon. Would this in any way contradict the given facts? (Think about the definition of lim sup)
 
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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