Prove that the sequence $3+\frac{1}{n^2}$ is Cauchy Question: Prove that the sequence defined by ${x_n}=3+\frac{1}{n^2}$ is Cauchy.
My attempt: Need to show that $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $m,n>N \implies|x_n-x_m|<\varepsilon.$
Fix $\epsilon>0$. (Edit: and let $m>n$.)
We have $|x_n-x_m|=|3+\frac{1}{n^2}-(3+\frac{1}{m^2})|$
$=|\frac{1}{n^2}-\frac{1}{m^2}|$
$=|\frac{m^2}{m^2n^2}-\frac{n^2}{m^2n^2}|$
$=|\frac{m^2-n^2}{m^2n^2}|$
$\le|\frac{m^2}{m^2n^2}|$
$=|\frac{1}{n^2}|=\frac{1}{n^2}.$
Then $\frac{1}{n^2}<\epsilon \implies n^2>\frac{1}{\epsilon}\implies n>\sqrt\frac{1}{\epsilon}$.
So choose $N=\lfloor\sqrt\frac{1}{\epsilon}\rfloor+1$ and the definition is satisfied. Q.E.D.
Is this correct? Thanks.

Just want to add my thought to this as you did most of it. Often times, you replace the minus $-$ by the plus $+$. So this means:
$\left|\dfrac{m^2-n^2}{m^2n^2}\right| \le \dfrac{m^2+n^2}{m^2n^2}= \dfrac{1}{m^2}+\dfrac{1}{n^2}\le \dfrac{1}{m}+\dfrac{1}{n} < \dfrac{1}{N}+\dfrac{1}{N}=\dfrac{2}{N}< \epsilon$, if $N > \dfrac{2}{\epsilon}$. Thus you can say: choose $N > \dfrac{2}{\epsilon}$ and for $m, n > N$ then you have $|x_m-x_n| < \epsilon$, proving it a Cauchy Sequence.
