Prove divergence of $\sin(n)$ using sub-sequence.

Question

Consider the sequence $a_n$ of the infinite decimal expansion of ${\pi}$. If I prove that the sequence $\sin(a_n)$ converges to $\sin({\pi})$ and similarly consider sequence $b_n$ of the infinite decimal expansion of ${e}$ , then $\sin(b_n)$ converges to $\sin({e})$ .

As both sub-sequence converge to different limits, does this imply the divergence of $\sin(n)$ for $n$ belonging to real numbers?

Actually, the set ${\sin n:n\in\Bbb N}$ is dense in $[-1,1]$. See here https://math.stackexchange.com/questions/3820747/infimum-supremum-of-the-sequence-sin-n/3820753#3820753 — Sumanta, Sep 29 '20 at 07:13
Thanks, I will check it out. So l guess my intuition is wrong for the proposed proof outline above . — Aristotle Stagiritis, Sep 29 '20 at 07:17
@Aristotle The part that's wrong is that it doesn't make sense to consider a sequence of integers $b_n$ that converge to $e$. However, you could certainly find a sequence $b_n$ for which $\sin(b_n)$ converges to $\sin \theta$ for some $\theta \neq \pi$. — Ben Grossmann, Sep 29 '20 at 08:18
@Ben Grossman, l meant the sequence (3.14,3.141,3.1415,.......) for $a_n$ and the sequence (2.7,2.71,2.718,2.7182,.........) for $b_n$ — Aristotle Stagiritis, Sep 29 '20 at 09:48
@AristotleStagiritis what Ben is saying is, $a_n$ and $b_n$ need to be integers in order for $\sin(a_n),\sin(b_n)$ to be subsequences of $\sin n$ — Calvin Khor, Sep 29 '20 at 11:15

score 1 · Accepted Answer · answered Sep 29 '20 at 10:53

Suppose that the sequence $(\sin n )$ converges to a finite limit $g.$ Then the subsequence $(\sin (2n) )$ also converges to $g. $ Hence the sequence $$(|\sin (2n) -\sin (2n +2)|)$$ converges to $0.$ Moreover the sequence $(|\cos n |)=(\sqrt{1-\sin^2 n})$ converges to $\sqrt{1-g^2 }.$ Therefore $$2\sin 1\sqrt{1-g^2}=2\sin1 \lim_{n\to\infty} \cos (2n +1) =\lim_{n\to\infty } |\sin (2n) -\sin (2n +2)| =0.$$ Therefore the sequence $(\cos (2n +1))$ converges to $0$ and hence the sequence the sequence $(|\cos n |) $converges to $0.$ Moreover the sequence $$(|\cos (2n+1) -\cos (2n +3)|)=(2\sin 1 \sin (2n+2) ) $$ is also convergent to $0$ and this implies that $(\sin n)$ is convergent to $0.$ But $$1=\sin^2 n +\cos^2 n \to 0^2 +0^2 =0.$$

Contradiction.

Thanks for providing the proof, looks nice. – Aristotle Stagiritis Sep 29 '20 at 13:51 — Aristotle Stagiritis, Sep 29 '20 at 13:51

score 1 · Answer 2 · answered Sep 29 '20 at 20:52

If we're interested in proving $\sin n$ diverges, note that $e^{in}$ belongs to $\{e^{it}: \pi/4<t< 3\pi/4\}$ for infinitely many $n,$ and belongs to $\{e^{it}: 5\pi/4<t< 7\pi/4\}$ for infinitely many $n.$ Thus $\sin n >1/\sqrt 2$ for infinitely many $n,$ and $\sin n <-1/\sqrt 2$ for infinitely many $n.$

Prove divergence of $\sin(n)$ using sub-sequence.

2 Answers2