Prove that if a_n is a sequence that contains no convergent subsequences, the lim abs(a_n) = infinity

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- Nov 3rd 2009, 12:47 PMJazz10a_n has no convergent subsequences => lima_n = infinity
Prove that if a_n is a sequence that contains no convergent subsequences, the lim abs(a_n) = infinity

- Nov 3rd 2009, 01:05 PMPlato
Can you prove that each sequence contains a monotone subsequence?

But any bounded monotone sequence converges.

So if you prove the first then you have solved this problem.

Here is a start. Define $\displaystyle S = \left\{ {k:\left( {\forall n > k} \right)\left[ {a_n > a_k } \right]} \right\}$

Two cases: $\displaystyle S$ is infinite or $\displaystyle S$ is finite.

Case one gives a increasing subsequence.

Case two gives a decreasing subsequence. - Nov 3rd 2009, 02:21 PMJazz10
I appreciate the help and see where you are going with it, however, can you clarify what you mean by each seqeuence? as in each subsequence has a monotone subsequence itself? not really sure how that ties in to your hint at the end

- Nov 3rd 2009, 03:04 PMPlato
Frankly, after I posted that I realized a simpler proof.

Define $\displaystyle \left( {\forall N \in \mathbb{Z}^ + } \right)\left[ {S_N = \left\{ {a_k :\left| {a_k } \right| \leqslant N} \right\}} \right]$.

What if some $\displaystyle S_J$ were infinite?

Would we have a convergent subsequence? WHY?

One real problems with help-sites such as this is simply that helpers have no idea what theorems the poster knows and therefore can use. - Nov 3rd 2009, 03:38 PMJazz10
ooo, nice, so using your setup, if we had an set Sj that were infinite, then that infinite set would be bounded and would therefor have an accumulation point, and thus we could find subsequence that was convergent. Since there are no convergent subsequences, there cannot exist a j that bounds any set, and therefore the sequence must diverge to inifity or negative infinity.

is that where you were going?