I first approached the problem using contradiction.
Suppose that |$a_n$| does not approach infinity.
Case 1: The sequence will converge to some L, and thus be bounded.
Case 2: The sequence will neither converge nor diverge, thus it will be bounded. (I'm having trouble explaining case 2)
For both cases, we can use the Bolzano-Weiestrass Theorem, thus |$a_n$| has a convergent subsequence.
(How can I show that $a_n$ will also have a convergent subsequence?)
This is a contradiction to the given statement, thus |$a_n$| will approach infinity.
This was how I approached the problem, but there's a few problems with it. Please let me know if there is a better way or how I can improve mine.
Thank you in advance!
OR
Again using contradiction, I can say that there exists an M > 0 for all N in the natural numbers, where there exists n > N such that we have |$a_n$| ≤ M.
Let n = $a_ni$. Then |$a_ni$| ≤ M. Thus -M < $a_ni$ < M. This sequence is bounded, thus, by the Bolzano-Weiestrass theorem, it has convergent subsequences. CONTRADICTION
Best Answer
Suppose by contradiction that $|a_n|$ doesn't diverge to $\infty$. This means that there exists $M>0$ such that, for all $n$, there exists $m>n$ with $|a_m|\le M$.
So, start with defining $n_0$ so that $|a_{n_0}|\le M$. Then take $n_1>n_0$ such that $|a_{n_1}|\le M$ and so on. More precisely, if $n_k$ has been defined, take $n_{k+1}>n_k$ so that $|a_{n_{k+1}}|\le M$.
The subsequence $(a_{n_k})$ has its values in $[-M,M]$, so it is bounded and therefore it has a convergent subsequence, which is also a convergent subsequence of the original sequence.