Revisited$^2$: Why is $(-n)^2$ divergent? How can it be shown rigorously? Why is $(-n)^2$ divergent? How is this proven? I've tried using the $\epsilon$ definition of convergence to come to a contradiction, but I don't know that using the definition is the way to go. I get that $n^2-s\leq \frac{1}{n}$. Not sure where to go from here. A hint would be nice.











Attempt1:

Say $\lim_{n\rightarrow\infty}=(-n)^2=s$. Then given an $\epsilon&gt;0$ we can find an $N\in\mathbb{N}$ so that $\lvert (-n)^2-s\rvert \leq \epsilon$ for every $n\geq N$. But if $n\geq N$, then we must have that
$$\lvert (-n)^2-s\rvert\leq \frac{1}{n}\leq\frac{1}{N},$$
but . . . hmm . . . I don't know that this is a fruitful approach. 







Attempt2:

Assuming that the negation of convergence is


  $$\text{$\exists\epsilon\leq 0$ s.t. $\forall N\in\mathbb{N}$ $n&gt;N$ so that $\lvert s_n-s\rvert \geq \epsilon$},$$


then if we let $\epsilon=0$, then $\lvert s_n-s\rvert \geq \epsilon\implies \lvert n^2-s\rvert \geq 0$, which means that $n^2\geq s$ and $s\geq n^2$, so $s=n^2$, which means $\lvert s_n-s\rvert \geq \epsilon$ holds as $0\geq 0$, right?
Easiest to use the negation of the definition of Cauchy sequence:

There exists $\epsilon &gt; 0$ such that for any $N$, there is some $n,m &gt; N$ such that $|x_n - x_m| &gt; \epsilon$.

Take time to parse this statement and it won't be too bad to prove what you want.
Note that $(-n)^2 = (-1)^2 n^2= n^2$. So, proving that $(-n)^2 \to \infty$ is the same as proving that $n^2 \to \infty$, which is an easy and well known proof.