Simple Limit Divergence I am working with a definition of a divergent limit as follows:

A sequence $\{a_n\}$ diverges to $-\infty$ if, given any number $M$, there is an $N$ so that $n \ge N$ implies that $a_n \le M$. 

The sequence I am considering is $a_n = -n^2$, which I thought would be pretty simple, but I keep running into a snag.
My Work: 
For a given $M$, we want to show that $n \ge N \Rightarrow -n^2 \le M$. So $n^2 \ge -M$. But here is where I run into trouble, because I can't take square roots from here. What should I do? 

An alternative (I think simpler) solution would be to note that $-n^2 \leq -n$ for $n \in \mathbb{Z}$ so taking $N=|M|$ suffices.
