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

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.

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

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

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

$$1/m+1/n;$$

Let $$\epsilon >0$$ be given.

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

For $$m\ge n \ge n_0:$$

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

• 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. – smalllearner May 8 '20 at 7:23
• 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:) – Peter Szilas May 8 '20 at 7:51
• Perfect, thank you very much – smalllearner May 8 '20 at 8:18
• smalllearner. Welcome. – Peter Szilas May 8 '20 at 8:22

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$$m,n\geqslant N\implies|x_m-x_n|\leqslant\frac1m+\frac1n\leqslant\frac2N<\varepsilon.$$

• 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. – smalllearner May 8 '20 at 7:23
• \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} – José Carlos Santos May 8 '20 at 7:49