# How to show that $\sin(n)$ does not converge ONLY by using Cauchy's criterion?

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?

• Am I allowed to do that? shouldn't m,n be natural numbers? – Buk Lau Dec 14 '18 at 14:32
• 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. – postmortes Dec 14 '18 at 14:35

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 $$|\sin(n_0) - \sin(n_1)| > 1$$.

• 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! – Buk Lau Dec 14 '18 at 15:09
• 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? – Buk Lau Dec 14 '18 at 15:26
• 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 – ODF Dec 14 '18 at 15:35

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.