$\mathbb{R}$ 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

  • $\begingroup$ As $n \to \infty$, $arctan(n) \to \frac{pi}{2}$. Does this help? $\endgroup$
    – Exit path
    Mar 14, 2016 at 20:10
  • $\begingroup$ i don't know, i want to find $n_0$ $\endgroup$
    – Vrouvrou
    Mar 14, 2016 at 20:15
  • $\begingroup$ Why? A convergent sequence is a Cauchy sequence. $\endgroup$ Mar 14, 2016 at 20:15
  • $\begingroup$ actually $(x_n=n)$ is not a convergent sequence, because if we suppose that is convergent to some x we have that $\lim|\arctan(n)-\arctan(x)|=0$ this means that $\arctan(x)=\pi/2$ then $x=\tan(\pi/2)$ contradiction @FriedrichPhilipp $\endgroup$
    – Vrouvrou
    Mar 14, 2016 at 20:20
  • $\begingroup$ Well observed. :o) But did I say that $(x_n)$ is convergent? ;) leibnewtz told you that $(\arctan(n))$ is convergent (in the usual metric). $\endgroup$ Mar 14, 2016 at 20:23

2 Answers 2


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.

  • $\begingroup$ i don't understand who is $n_0$ ? $\endgroup$
    – Vrouvrou
    Mar 14, 2016 at 20:22
  • $\begingroup$ Exactly what is written taking the definition of a sequence converging to a limit. $\endgroup$ Mar 14, 2016 at 20:23
  • $\begingroup$ in previous example we take $n_0=[...]+1$ $\endgroup$
    – Vrouvrou
    Mar 14, 2016 at 20:25
  • $\begingroup$ @Vrouvrou Why are you so stubborn? ;) It is not important which value $n_0$ has. It has to exist. That's the point. Here, we get its existence from the convergence of $(\arctan(n))$. $\endgroup$ Mar 14, 2016 at 20:27
  • $\begingroup$ thank you, please in general it is right to say that $(x_n)$ is Cauchy because $d(x_p,x_q)\rightarrow 0, p,q\rightarrow+\infty$ ? $\endgroup$
    – Vrouvrou
    Mar 14, 2016 at 20:33

$\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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.