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

Question

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?

Answer 1

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

Answer 2

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.