Can't do this supremum question

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Homework Statement



here's the picture and it's the second part of question 5:

http://imgur.com/ybSW4v4

Homework Equations



N/A

The Attempt at a Solution



so by intuition, I suspect that b = sup{a_n: n is in the natural numbers}

If we can show that, then it will follow from the first part of question 5 that what we are trying to prove does hold.

Ok well the problem is this. I'm given the definition of Least Upper Bound:

1) b is an upperbound
2) if y is an upperbound of the set, then (y > b) or (y = b).

I'm not sure how to prove 2) at all.. I need major help =/ Please..
 
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The question is: $$\text{Q5. }\text{Suppose that }a_n\leq b\text{ for all } n \text{ and that } a=\lim_{n\rightarrow \infty} a_n \text{ exists. }\\
\text{5.1 Show that }a\leq b\text{. }\\
\text{5.2 Conclude that } a\leq \sup_n a_n = \sup_n \{ a_n : n\in \mathbb{N} \}$$
... hmmm, how to describe it...

if A={1,2,3}, then sup[A] = what?,
if an ≤ b, what is b?
Is b supposed to be a single number or a set of them?
What does it mean to say "## a=\lim_{n\rightarrow \infty} a_n## exists"?
 
Oh crap... For whatever reason, i parsed "b" to be the number such that it is greater than AND equal to an. Thank you. So b is a set of numbers that satisfy an≤ b right?
 
I'm working on this same question. I believe b is supposed to be a real number so that the sequence is bounded.
 
Zondrina said:
I'm working on this same question. I believe b is supposed to be a real number so that the sequence is bounded.

I've posted a hint in Zondrina's post; you can check it out there: https://www.physicsforums.com/showthread.php?p=4364873#post4364873

Basic idea: consider two subsequences of your sequence a_n, one that is decreasing, and one that is increasing. Show that if you have an decreasing sequence, then your inequality is satisfied. Show that this also holds for increasing sequences. Conclude that this inequality is valid for your original sequence.
 
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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