Real Analysis: Sequences and Series

sprstph14
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Suppose that ak is a decreasing sequence and (ak) approaches 0. Prove that for every k in the natural numbers, ak is greater than or equal to 0.

I was thinking I should assume the sequence is bounded below by 0 and do a proof by contradiction.

Any suggestions?
 
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sprstph14 said:
Suppose that ak is a decreasing sequence and (ak) approaches 0. Prove that for every k in the natural numbers, ak is greater than or equal to 0.

I was thinking I should assume the sequence is bounded below by 0 and do a proof by contradiction.

Any suggestions?

If you assume the sequence is bounded below by 0 you are assuming what you are trying to prove.

Just try an indirect argument. Assume some ak < 0. You should be able to get an easy contradiction.
 
thanks. I guess I just really don't know what I'm doing.
 
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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