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Prove that the sequence $x_n = 1 +\frac{ sin (n+ \pi) }{n} $ is a cauchy sequence using the definition:

$$\forall \epsilon>0 \exists N\in\mathbb{N}: n,m\ge N\implies |x_n-x_m|<\epsilon.$$

I have tried to prove: $ | \frac{ n-sin(n)}{n} - \frac{ m -sin(m)}{m} | \leq \epsilon$. The triangle inequality did not work for me and I don't know how to prove it with the provided definition.

I could argue that the sequence converges to 1 and is therefore cauchy. However, I need to prove this with the definition.

Please help me.

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  • $\begingroup$ $\frac {n-\sin(n)}{n} = 1 - \frac {\sin n}{n}$ With that simpification the problem gets easier. $\endgroup$ – Doug M May 7 '20 at 20:26
  • $\begingroup$ @JonathanZsupportsMonicaC Thank you very much, it was a typo. $\endgroup$ – smalllearner May 7 '20 at 20:26
  • $\begingroup$ $$\sin(n+\pi) = \sin n \cos \pi + \sin \pi \cos n = -\sin n$$ so this is really $$x_n = 1 - \dfrac{\sin n}{n}$$ isn't it? $\endgroup$ – InterstellarProbe May 7 '20 at 20:26
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$|\frac{\sin (π+m)}{m}-\frac{\sin (π+n)}{n}|\le$

$|\frac{\sin (π+m)}{m}| +|\frac{\sin (π+n)}{n}|\le$

$1/m+1/n;$

Let $\epsilon >0$ be given.

Choose $n_0 > 2/\epsilon$ (Archimedean principle).

For $m\ge n \ge n_0:$

$|\frac{\sin (π+m)}{m}-\frac{\sin (π+n)}{n}|\le 1/m+1/n \le 2/n \le 2/n_0 <\epsilon.$

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  • $\begingroup$ Hello thank you very much. Could you please tell me why it's $\leq 1/m + 1/n$? I'm learning on my own and I don't know how we got there. Thanks again. $\endgroup$ – smalllearner May 8 '20 at 7:23
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    $\begingroup$ smalllearner. Triangle inequality:$ |a +b| \le |a|+|b|$, now: $|\sin(pi +m)/m+\sin (pi+n)/n| \le |\sin(pi+m)/m|+|\sin (pi +n)/n| =|\sin (..)|/m+|\sin (..)|/n\le 1/m+1/n$ since $|\sin (pi+n)| \le 1$, same for m index.Ok, if not come again:) $\endgroup$ – Peter Szilas May 8 '20 at 7:51
  • $\begingroup$ Perfect, thank you very much $\endgroup$ – smalllearner May 8 '20 at 8:18
  • $\begingroup$ smalllearner. Welcome. $\endgroup$ – Peter Szilas May 8 '20 at 8:22
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Note that\begin{align}|x_m-x_n|&=\left|1+\frac{\sin(m+\pi)}m-\left(1+\frac{\sin(n+\pi)}n\right)\right|\\&=\left|\frac{\sin(m+\pi)}m-\frac{\sin(n+\pi)}n\right|\\&\leqslant\frac1m+\frac1n.\end{align}So, given $\varepsilon>0$, take $N\in\Bbb N$ such that $\frac1N<\frac\varepsilon2$ and then$$m,n\geqslant N\implies|x_m-x_n|\leqslant\frac1m+\frac1n\leqslant\frac2N<\varepsilon.$$

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  • $\begingroup$ Hello thank you very much. Could you please tell me why it's $\leq 1/m + 1/n$? I'm learning on my own and I don't know how we got there. Thanks again. $\endgroup$ – smalllearner May 8 '20 at 7:23
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    $\begingroup$ \begin{align}|x_m-x_n|&=\left|\frac{\sin(m+\pi)}m-\sin{(n+\pi)}n\right|\\&\leqslant\left|\frac{\sin(m+\pi)}m\right|+\left|\frac{\sin(n+\pi)}n\right|\text{ (triangle inequality)}\\&\leqslant\frac1m+\frac1n\end{align} $\endgroup$ – José Carlos Santos May 8 '20 at 7:49

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