Prove: If $|a_n|$ doesn't converge to $\infty$ then $a_n$ must have a finite partial limit. Prove: If $|a_n|$ doesn't converges to $\infty$ then $a_n$ must have a finite partial  limit.  
My thoughts:
if $|a_n|$ doesn't converges to $\infty$ there must be two other posibilities:  


*

*$|a_n|$ converges to a finite number,  $L$

*$|a_n|$ doesn't have a limit at all (neither finite, nor $\infty$)


for option #1, we can infer $a_n$ is bounded by $L$ and has a converges parital limit (by BW Theorem).
EDIT:
The correct demand is to prove $a_n$ has a finite partial limit.
for option #1 I proved it using BW Theorem.
Now I need to prove $a_n$ has a partial finite limit for option #2
 A: If it doesn't converge to infinity then it must have a bounded subsequence(if it had no bounded subsequence, it would converge to infinity), this bounded subsequence then has a convergent subsequence, which itself is a subsequence of the original sequence.
A: We have $\lim_{n \to \infty} a_n = L$ if and only if $\liminf_{n\to \infty} a_n = \limsup_{n\to \infty} a_n =L$. Here $L$ can be finite or infinite. If $L=\infty$, we have something stronger and conclude $\lim_{n \to \infty} a_n = L$ if and only if $\liminf_{n \to \infty} a_n = L$. It is enough to use facts about the liminf.
A: This is basically the same answer as mathematician's:
For each integer $m$ we define 
$$A_m:= \{ n | a_n \in [m,m+1) \} \,.$$
If all $A_m$ are finite, then it is easy to prove that 
$$\lim_n |a_n| =\infty \,.$$
Indeed, for each positive integer $M$ we have
$$|a_n| >M$$if and only if 
$n$ is not in the finite set  $\cup_{k=-M}^m A_k$. In particular, with $j$ being the largest element in this set we have
$$n > j \Rightarrow |a_n| >M \,.$$
Thus, some $A_m$ must be infinite. This $A_m$ defines a bounded subsequence, and hence we can find a subsequence of this which is convergent.
A: Let $a_n=(-1)^n$. Then $|a_n|$ converges to $1$ but $a_n$ does not converge.
