Determine using only the definition if it is a Cauchy sequence $(2n^2+1)/n^2$
Where did I go wrong?
This is what I have so far:
$$\frac{2m^2+1}{m^2}-\frac{2n^2+1}{n^2}= \frac{n^2(2m^2+1)}{m^2n^2}-\frac{(2n^2+1)m^2}{m^2n^2}= \frac{n^2-m^2}{m^2 n^2}$$
$$|x_m-x_n|=\frac{|n^2-m^2|}{|m^2n^2|}≤ \frac{m^2}{m^2n^2}<\frac{1}{n^2}≤\frac{ 1}{N^2}<ε $$
A sequence is cauchy if for any $\epsilon > 0$ then is an $N$ so that for any $n,m \ge N$ then $|x_n -x_m| < \epsilon$.
Here for any $n,m$ we have $|x_n - x_m|< \frac 1{n^2}$ so for any $N$ if $n,m \ge N$ we will have $|x_n - x_m| < \frac 1{N^2}$.
Now given any $\epsilon$ if we can find an $N$ so that $\frac 1{N^2} < \epsilon$ that will prove that the sequence is Cauchy. For any $\epsilon > 0$ can we find an $N$ so that $\frac 1{N^2} < \epsilon$?
Well... yes... $\frac 1{N^2} < \epsilon \iff 0< \frac 1{\epsilon} < N^2 \iff \sqrt{\frac 1{\epsilon}} < |N|$.
So for any $\epsilon > 0$ if we let $N > \sqrt{\frac 1{\epsilon}}$ then for any $n,m \ge N$ we have $|x_n - x_n| < \frac 1{N^2} < \epsilon$. So it's Cauchy.