Limit of sequence (by definiton) I have this assignment:
$$\lim_{n\to \infty }\frac{(-1)^n-3}{n^2}=0$$
I have to prove that limit by definition, but I am stuck with 
$$|a_n - 0|< \epsilon$$
Dont know how to deal with that absolute value.
The solution is
$$n_0=\left[\frac{2}{\sqrt\epsilon}\right]+1$$
 A: Given $\epsilon>0 $, we should find $N$ such that if $n\ge N$ then $$|\frac{(-1)^n-3}{n^2}|<\epsilon$$
but
$$|\frac{(-1)^n-3}{n^2}|\le |\frac{(-1)^n}{n^2}|+|\frac{3}{n^2}|\le \frac{4}{n^2}$$
thus it is sufficient to find one $N$ such that,
$$n\ge N \;\; \implies \;\; \frac{4}{n^2}<\epsilon$$
or
$$n\ge N \;\;  \implies \;\; n^2 > \frac{4}{\epsilon}$$
or
$$n\ge N \;\; \implies \;\; n > \frac{2}{\sqrt{\epsilon}}$$
So, each  $N$  satisfying $N>\frac{2}{\sqrt{\epsilon}}$ will work.
the smallest of these $N$ is
$$\lfloor \frac{2}{\sqrt{\epsilon}} \rfloor   +1$$
A: We have that
$$\frac{-2}{n^2}\le\frac{(-1)^n-3}{n^2}\le \frac{-4}{n^2} $$
and then
$$\frac{2}{n^2}\le\left|\frac{(-1)^n-3}{n^2}\right|\le \frac{4}{n^2} $$
therefore
$$\left|\frac{(-1)^n-3}{n^2}\right|\le \frac{4}{n^2} <\epsilon \implies n^2 > \frac{4}{\epsilon} \implies n>\frac{2}{\sqrt\epsilon}$$
therefore it suffices to take
$$n_0=\overbrace{\color{blue}{\left[\frac{2}{\sqrt\epsilon}\right]}}^{\color{red}{\text{integer part}}}+1$$
A: Notice, that:
$$\frac{(-1)^n-3}{n^2} \geq \frac{-4}{n^2}$$
And is always negative. So for every $\epsilon$ you can find $N$ ($=\frac{2}{\sqrt{\epsilon}})$, that for every $n>N$ the following holds: $$|\frac{(-1)^n-3}{n^2}| \leq \epsilon$$
