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