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 $\epsilon>0$ 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

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

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?
 A: Easiest to use the negation of the definition of Cauchy sequence:
There exists $\epsilon > 0$ such that for any $N$, there is some $n,m > N$ such that $|x_n - x_m| > \epsilon$.
Take time to parse this statement and it won't be too bad to prove what you want.
A: 
Claim Every convergent sequence is bounded.

Proof Suppose that $x_n\to x$. Given $\epsilon =1$, there is $N$ such that if $n\geqslant N$ then $$|x_n|-|x|<|x-x_n|<1$$ so when $n\geqslant N$ $$|x_n|<1+|x|$$
Then for each $n$ we have $$|x_n|\leq \max\{|x_1|,\ldots,|x_{N-1}|,|x|+1\}$$
Observation The sequence $a_n=(-n)^2$ is not bounded.
A: Working from the definition ,$\forall L \in \mathbb{R} \ \exists \epsilon > 0, \forall N \in \mathbb{N} \exists \ n > N : |n^2 - L| \geq \epsilon$.
If L is negative or zero, it is easy to see that $n^2 + L$ that $\forall n \geq 1 \ n^2 + L > L.$  Choose $\epsilon = |L|, $ so $|n^2 - L| = n^2 + L > |L| = \epsilon.$
If L is positive, and $N \leq \lfloor\sqrt{L}\rfloor$ and N $\neq \sqrt{L}$, choose $\epsilon = L$. Then: 
$\forall n \geq \lceil{\sqrt{2L}}\rceil, \ |\lceil{\sqrt{2L}}\rceil^2 - L| \geq |\sqrt{2L}^2 - L| = |L| = L = \epsilon.$
If L is positive and $N = \sqrt{L}$ or $N \geq \lceil\sqrt{L}\rceil$, then $\forall n \geq N \ |(n+1)^2 - L| > |n^2 -L|.$  
Choose $\epsilon = |(N+1)^2 -L|$ so $\forall n \geq N + 2$:
$|n^2 - L| > |(N+1)^2 -L| = \epsilon$.
A: Given $N > 0$, we can choose any number $N'$ such that
$\quad n > N'\ \Longrightarrow\ -n^{2} < N$.
For a given $N \leq 0\quad$,
$\forall\ n > \left\lfloor\sqrt{\vphantom{\large A}-N\,} + 1\right\rfloor$
it's true that $-n^{2} < N$. 
A: 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.
