Prove that the sequence $(-1)^n (-n)^2/(9n^2)$ does not converge to a specific number

Question

Prove that $\{(-1)^n (-n)^2/(9n^2)\}$ does not converge to $\frac{1}{9}$ or $\frac{-1}{9}$, I've already chosen an epsilon whats the next step. DO NOT SOLVE ALL THE WAY! Do I chose an $N$?

Answer 1

$$\frac{(-1)^{(-n)^2}}{9n^2}=\frac{(-1)^{n^2}}{9n^2}\to 0\text{ if }n\to\infty $$ Moreover, the limit is unique, indeed, if a sequence $(x_n)$ converge to $\ell$ and $\ell'$, then if $\varepsilon>0$, $$|\ell-\ell'|\leq|x_n-\ell|+|x_n-\ell'|<\varepsilon$$ if $n>N$ for some $N\in\mathbb N$, and so $\ell=\ell'$, what terminate the proof.

Answer 2

No, you don't get to choose an $N$. The negation of "$x_n\to a$" is: there is $\epsilon>0$ such that for every $N$ there is $n>N$ for which $|x_n-a|\ge \epsilon$. This means:

1. you get to choose $\epsilon$
2. you have no control on $N$
3. you have to come up with $n$ such that $n>N$ and $|x_n-a|\ge \epsilon$.

It seems that one of $n=N+1$ and $n=N+2$ ought to work, if your choice of $\epsilon$ was good.