TITLE: Prove that every unbounded sequence contains a monotone subsequence that diverges to inifnity.
QUESTION [7 upvotes]: I think I have the basic framework for this proof, but I am having trouble putting everything together in a convincing way.  
I plan on showing that by the definition, an unbounded sequence $(a_n)$ has infinitely many $n$ such that $a_n>M$ where $M>0$, for some natural number $n$. I'm not sure how it is implied that there exists some $a_{n_k}$ such that $a_{n_k}>M$ for some natural number $k$.  Let me know if I'm on the right track, and please help me string these ideas together!

REPLY [3 votes]: I believe it's because you'd like to guarantee the convergence of $a_{n_k}$ to $\infty$. Actually, you could use the terms of any divergent sequence $b_k$ to do that comparation. 
For example, you could chose $n_k$ to guarantee that $a_{n_k}>k^2$. The reasoning would be the same, because $b_k = k^2 \rightarrow \infty$ when $k \rightarrow \infty $