# 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?

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If $-M$ is negative, then $n^2\geq -M$ automatically, otherwise you can take the square root. – nullUser Aug 13 '12 at 13:35
Ah thank you both @nullUser – Moderat Aug 13 '12 at 14:20

If $M>0$ inequality $n^2\geq -M$ always holds. So you can take any $N$ you want.
If $M\leq 0$, then $n\geq\sqrt{-M}$, and you can take $N=\lfloor\sqrt{-M}\rfloor +1$.
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.