How to prove that is a Cauchy sequence

Question

$\mathbb{R}$ is endowed with the metric $d(x,y)=|\arctan(x)-\arctan(y)|$ ,

i want to prove that $(x_n=n)$ is a Cauchy sequence, i see that $\lim_{p,q\rightarrow\infty} d(x_p,x_q)=0$ then $(x_n)$ is a cauchy sequence, but if i want to find $n_0$ , how to do for $|\arctan(p)-\arctan(q)|<\varepsilon$

i mean how to prove using the definition:$$\forall\varepsilon>0, \exists n_0\in \mathbb{N}, \forall p,q\in \mathbb{N}, p>q\geq n_0\Rightarrow d(x_p,x_q)<\varepsilon$$ we have

$|\arctan(p)-\arctan(q)|<|\arctan(p)|+|\arctan(q)|\leq 2 |\arctan(p)|<\varepsilon$

how to find $n_0$ ?

Thank you

Answer 1

As $\arctan$ is increasing, for $0 < n < m$, you have $$0 < \arctan m - \arctan n < \frac{\pi}{2}-\arctan n$$

As $$\lim\limits_{n \to \infty} \arctan n =\frac{\pi}{2},$$ you can take $n_0$ such that for $n \ge n_0$ $$0 <\frac{\pi}{2}-\arctan n <\epsilon$$ For $n_0 \le n<m$ you'll get

$$0 < \arctan m - \arctan n < \frac{\pi}{2}-\arctan n < \epsilon$$ as desired.

Answer 2

$\arctan(n)-\arctan(m) =\arctan(\frac{n-m}{1+nm}) $ so if $\min(n, m) \ge v$ and $n \ge m$, $|\arctan(n)-\arctan(m)| =|\arctan(\frac{n-m}{1+nm})| \le|\arctan(\frac{n-v}{1+nv})| < \frac1{v} $.

Therefore, if $n, m > \frac1{\epsilon} $, then $|\arctan(n)-\arctan(m)| < \epsilon $, so the sequence is Cauchy.