# Unbounded sequences

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#1
Hey need some help with proving this statement;
"A sequence (an) is not bounded above if and only if there is a sub-sequence (akn) such that this sub-sequence tends to infinity."

I've proved that if the sequence is not bounded above implies that there is a sub-sequence which tends to infinity but I can't see how to construct a formal proof for the other direction?

I know that if a sequence converges, then every subsequence converges to the same limit. Hence if you have a divergent subsequence, the sequence cannot converge, but don't know if this is sufficient.

0
5 years ago
#2
(Original post by Kim-Jong-Illest)
Hey need some help with proving this statement;
"A sequence (an) is not bounded above if and only if there is a sub-sequence (akn) such that this sub-sequence tends to infinity."

I've proved that if the sequence is not bounded above implies that there is a sub-sequence which tends to infinity but I can't see how to construct a formal proof for the other direction?
Write down what it means to say that a subsequence tends to infinity. It should be obvious how to show this means the subseqeunce cannot be bounded above, and so neither can the original sequence.

I know that if a sequence converges, then every subsequence converges to the same limit. Hence if you have a divergent subsequence, the sequence cannot converge, but don't know if this is sufficient.
I don't see how this is relevant (and to be honest, the fact that you think it is makes me suspect you are rather confused about this topic).
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#3
(Original post by DFranklin)
Write down what it means to say that a subsequence tends to infinity. It should be obvious how to show this means the subseqeunce cannot be bounded above, and so neither can the original sequence.

I don't see how this is relevant (and to be honest, the fact that you think it is makes me suspect you are rather confused about this topic).
I am pretty confused, need to learn the definitions more rigorously.

If (ank) is divergent to +infinity then for all a in the real numbers, there exists >0 such that for all N in the natural numbers, there exists a k>=N where | (ank) - a | > . Since this holds for all N it means that the sequence is not bounded from above and therefore the mother sequence cannot be either.

Is this correct? If so I feel like my conclusion should be more formal but not sure how to express it.

0
5 years ago
#4
if such that n > N .

Now suppose a_n is bounded above, let M be the upper bound and compare with the sentence above...
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