Question: How to prove that the sequence $\{a\_n\}$ has no convergent subsequence if and only if $|a\_n|$ approaches infinity?
The forward direction is obvious, how to prove the other direction?

Answer: 
Assume $\lim\_{n\to\infty}|a\_n|=\infty$ and consider an $\alpha\in{\mathbb R}$. There is an $n\_0$ such that $|a\_n|>|\alpha|+1$ for all $n>n\_0$. Therefore $|a\_n-\alpha|\geq|a\_n|-|\alpha|>1$ for all $n>n\_0$, and this implies that the sequence $(a\_n)\_{n\geq1}$ cannot have a subsequence converging to $\alpha$.

For the converse, assume that $\lim\_{n\to\infty}|a\_n|=\infty$ does not hold. This means that there is an $M>0$ such that there are arbitrary large $n$ with $|a\_n|\leq M$. This can be exploited as follows: Put $n\_0:=0$. Then for each $k\geq1$ we can find and $n\_k>n\_{k-1}$ such that $|a\_{n\_k}|\leq M$. The subsequence $(a\_k')\_{k\geq1}$ defined by $a\_k':=a\_{n\_k}$ $\>(k\geq1)$ is bounded, therefore it has a subsequence $(a\_l'')\_{l\geq1}$ converging to some $\alpha\in{\mathbb R}$. This sequence $(a\_l'')\_{l\geq1}$ can be considered as a subsequence of the originally given sequence $(a\_n)\_{n\geq1}$.
**Hint** If this result not true then [weierstrass sequence theorem](http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem) is false. 

**Up date** Imitate this [proof](http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem#Proof) of [weierstrass sequence theorem](http://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem) (with necessary adaptations of course) to get the proof of this fact. What I mean is that the same idea of this proof of weierstrass sequence theorem can be used to prove this result.