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$\mathbb{R}$ 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

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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
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$\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.

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