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