Real Analysis - A sequence that has no convergent subsequence

Use these two facts:

• Every sequence has a monotone subsequence.
• Every bounded monotone sequence converges.

First show that, if $|a_n|$ doesn't approach $\infty$, then $a_n$ has a bounded subsequence; then invoke the fact that a bounded sequence of real numbers has a convergent subsequence.

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 $(a_n)_{n\geq1}$.