$\mathbb{R}$ 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? $\endgroup$ – leibnewtz Mar 14 '16 at 20:10
  • $\begingroup$ i don't know, i want to find $n_0$ $\endgroup$ – Vrouvrou Mar 14 '16 at 20:15
  • $\begingroup$ Why? A convergent sequence is a Cauchy sequence. $\endgroup$ – Friedrich Philipp Mar 14 '16 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 '16 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). $\endgroup$ – Friedrich Philipp Mar 14 '16 at 20:23

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 '16 at 20:22
  • $\begingroup$ Exactly what is written taking the definition of a sequence converging to a limit. $\endgroup$ – mathcounterexamples.net Mar 14 '16 at 20:23
  • $\begingroup$ in previous example we take $n_0=[...]+1$ $\endgroup$ – Vrouvrou Mar 14 '16 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 $(\arctan(n))$. $\endgroup$ – Friedrich Philipp Mar 14 '16 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 '16 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.