How to show that $\sin(n)$ does not converge?

Question

I am supposed to show that $\sin(n)$ does not converge by constructing two subsequences: one subsequence contains terms of $\sin(n)$ that are between $1/2$ and $1$, and the other subsequence contains terms of $\sin(n)$ that are between $-1/2$ and $-1$. But how can I do this? Which $k(n)$--for subsequence $x_{k(n)}$--should I choose?

Answer 1

Consider $I_k=[2k\pi+\frac{\pi}{6},2k\pi+\frac{5\pi}{6}]$, the length is $\frac{2\pi}{3}>1$, then there must $\exists n_k\in \mathbb{N}$ s.t $n_k\in I_k$. (Here we use the property that $\forall a>0, \exists n \in \mathbb{N}$,s.t. $a\leq n<a+1$.) Thus for each $k$, we find such $n_k$, s.t $\sin n_k>\frac{1}{2}$.

Similarly for $I_k^{'}=[2k\pi+\frac{7\pi}{6},2k\pi+\frac{11\pi}{6}]$. We get a subsequence $\{\sin n_k^{'}\}<-\frac{1}{2}$. Hence we can find find two subsequences don't converge to the same value. The whole sequence can't converge.

Answer 2

I believe you don't need to split sequence into two subsequences here, as an expression for finding sign of $\sin(n)$ doesn't look elegant for $n \in \mathbb N$. I think it's better to prove the problem with the help of Cauchy's convergence test:
Let $a_n = \sin n$ converge to $L$. Then $\forall \epsilon > 0, \exists N(\epsilon) > 0, \forall n,m > N(\epsilon) : |a_n - a_m| < \epsilon$.
Let $m = n + 1$. Then $|a_n - a_m| = |\sin n - \sin (n+1)| = |2\sin\frac 12 * \cos\frac{2n+1}{2}| = 2\sin\frac 12|\cos \frac {2n+1}{2}|$. This expression should be less than $\epsilon, \forall \epsilon > 0$ starting from some $N(\epsilon)$ but cosine always take values from $[-1,1]$. We came to a contradiction $\Rightarrow (a_n)$ does not converge.