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I know this question has been asked before... I went through all of the questions of this sort and none of them had an answer using Cauchy's criterion.

I know that $\sin(n)$ does not converge and I know how to show it in different ways (sub-sequences and unity of the limit), but I'm stuck with Cauchy... I can't figure it out.

I have to show that: $\exists \epsilon>0$ such that $\forall N\in\mathbb N, \exists m,n > N$ such that $|\sin(m)−\sin(n)|>\epsilon$.

How do I find $\epsilon$ and $m,n$ that will do it?

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  • $\begingroup$ Am I allowed to do that? shouldn't m,n be natural numbers? $\endgroup$ – Buk Lau Dec 14 '18 at 14:32
  • $\begingroup$ I'll delete my comment then as it's not relevant :) Since $\sin$ is periodic with an irrational period I'm not sure you can have an easy formula as answer. $\endgroup$ – postmortes Dec 14 '18 at 14:35
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Pick $\varepsilon = 1$. Let $N \in \mathbb{N}$. Let $k$ be such that $2k\pi > N$. Let $a_0 = 2k\pi + \frac{\pi}{6}$, $a_1 = 2k\pi + \frac{5\pi}{6}$, $b_0 = 2k\pi + \frac{7\pi}{6}$ and $b_1 = 2k\pi + \frac{11\pi}{6}$.

Since $a_1 - a_0 > 1$, there must exist an integer $n_0\in(a_0,a_1)$. Similarly there exists an integer $n_1 \in (b_0,b_1)$. We know that $\sin(a_0) = \sin(a_1) = \frac{1}{2}$, so by looking at the graph of $\sin(x)$, we see $\sin(n_0) > \frac{1}{2}$. Similarly, $\sin(n_1) < -\frac{1}{2}$.

Therefore, $n_0,n_1 > N$ and $|\sin(n_0) - \sin(n_1)| > 1$.

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  • $\begingroup$ I really like the way you do it here, I didn't think of that! I however think that I've managed to solve it differently, I'll post my answer in a bit for verification if that's possible. Thank you! $\endgroup$ – Buk Lau Dec 14 '18 at 15:09
  • $\begingroup$ Never mind.. I found a mistake in my proof. I'll stick with your way, but do I have to show what those integers n1,n0 are? $\endgroup$ – Buk Lau Dec 14 '18 at 15:26
  • $\begingroup$ No - it's enough to know that they exist. Something like $n_0 = \lceil a_0 \rceil$ and $n_1 = \lfloor b_1 \rfloor$ should work though $\endgroup$ – ODF Dec 14 '18 at 15:35
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Here's a strategy:

(1) Prove that if $\forall N, \exists m,n >N: \sin(n)> 10^{-10}, \sin(m) < -10^{-10}$, then the series diverges.

(2) Prove that $\forall N \exists n,m: \sin(n)>0, \sin(m)<0$

(3) Prove that if $|\sin(n)|<10^{-10}$, then $n=\pi k+r$ for integer $k$ and $r<10^{-9}$

(4) Use (3) to prove that if $|\sin(k)|<10^{-10}$, then $ \exists n,m >k : \sin(n)>10^{-10}, \sin(m)<-10^{-10}$

That is, we can always find a number $n$ with a positive sine. If the sine is less than $10^{-10}$, then $n$ mod $\pi$ must be less than $10^{-9}$, so $(n+1)$ mod $2pi$ must be greater than $10^{-9}$, so its sine is greater than $10^{-10}$. Similarly, we can find $m$ with sine less than $-10^{-10}$, and so taking $\epsilon=2*10^{-10}$, we find that the series is not Cauchy.

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