Is a Monotone Sequence with a Bounded Subsequence Always Bounded?

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Homework Statement



prove that a monotone sequence which has a bounded subsequence is bounded

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The Attempt at a Solution

 
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did you even try to solve this?
 
Of course

I tried using the def. of a monotone sequence to show that the subsequence was monotone and bounded hence it converged to some number and then tried to prove that the sequence was convergent thus it was bounded
 
The problem is of course that there could be a lot of convergent subsequences. Try taking the sup of the limits over all of the convergent subsequences. Claim: the sequence converges to this sup.

I think this should do it.
 

Thread 'Finding the nth roots of a complex number'

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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