a sequence sin(n) diverges

[gefest_md]

This is a problem from a school textbook.
How to prove that the sequence $<\sin n>_{n\geqslant1}$ diverges? I have been thinking that among the numbers $\sin n,\ \sin(n+1)\ \sin(n+2)\ \sin(n+3)$ there is some negative. Is there any simple proof of it. If first three are positive then fourth is negative? Also I do not know what to say if zero is not the limit.

[gefest_md]

I have read some part about the sequence $<\sin n^\circ>$ that if it had a limit than it would have been that $\mid\sin n_1-\sin n_2\mid<\frac12$ And it follows ... but in our sequence there are terms with difference between them larger than $\frac12$. This explanation has not satisfied me.

[gefest_md]

After a little correction I arrive to here: If $\sin n>0$ and $\sin(n+2)>0$, then $\sin(n+4)<0$.
Proof. Suppose $\sin n>0$ and $\sin(n+2)>0$. Then $0<\sin(n+2)=\sin n\cos 2+\cos n\sin 2$. Then $\cos n>0$. It follows $\sin(n+4)=\sin n\cos 4+\cos n\sin 4<0$.

So the limit can be neither positive nor negative. Zero case pending.

[Joker_vD]

gefest_md в сообщении #530217 писал(а):

So the limit can be neither positive nor negative.

Why is it so? What if $\sin n>0$, $\sin (n+2)>0$ holds only for a finite number of $n$'s? Then you can throw away a finiite number of members and end with a possibly converging sequence.

[gefest_md]

yes it was my wrong idea. but i have found some solution. time out

[gefest_md]

Is it true that from this:

gefest_md в сообщении #530217 писал(а):

If $\sin n>0$ and $\sin(n+2)>0$, then $\sin(n+4)<0$.

follows $\forall N\in\mathbb{N}\exists n\in\mathbb{N}(n\geqslant N\,\&\,\sin n<0)$?
If so then I think sequence can not have positive limit.

[gefest_md]

I am not sure for the moment that I can solve initial problem, but I am going to try to show that if it is true that

If $\sin n>0$ and $\sin(n+2)>0$, then $\sin(n+4)<0$

then

$\forall N\in\mathbb{N}\exists n\in\mathbb{N}(n\geqslant N\,\&\,\sin n<0)$

Suppose $N$ is arbitrary. Then I examine a structure of cases

(I) $\sin N<0$. Then there is an $n=N$ such that $n\geqslant N$ and $\sin n<0$.

(II) $\sin N>0$. Then I add two more subcases:

(1) $\sin(N+2)<0$. Then there is an $n=N+2$ such that $n\geqslant N$ and $\sin n<0$.

(2) $\sin(N+2)>0$. Then by the proposition from above $\sin(N+4)<0$. Thus there is an $n=N+4$ such that $n\geqslant N$ and $\sin n<0$.

Now I do not know what follows. I just stoped.

[Nimza]

Pass to the limit in $\sin(n+1) = \sin(n)\cos(1)+\cos(n)\sin(1)$.

[Maslov]

Assume that $\{\sin n\}$ converges, then it's fundamental, then

$(\forall \varepsilon > 0)( \exists N > 0)(\forall n > N) |\sin (n+2) - \sin n| < \varepsilon$

$\text{Now, try to find} \lim \limits_{n\to\infty} \cos n ~~\text{and} \lim \limits_{n\to\infty} \sin n$

[gefest_md]

Nimza в сообщении #530540 писал(а):

Pass to the limit in $\sin(n+1) = \sin(n)\cos(1)+\cos(n)\sin(1)$.

I understand this as a solution of zero limit.

Maslov в сообщении #530543 писал(а):

Assume that $\{\sin n\}$ converges, then it's fundamental, then

$(\forall \varepsilon > 0)( \exists N > 0)(\forall n > N) |\sin (n+2) - \sin n| < \varepsilon$

$\text{Now, try to find} \lim \limits_{n\to\infty} \cos n ~~\text{and} \lim \limits_{n\to\infty} \sin n$

Cauchy sequence intrigues me. For the rest I can not apply this hint. But I said earlier I found a solution from some other place. Here it is

$$\lim_{n\to\infty}\sin n=a$$

Then obviously

$$\lim_{n\to\infty}\sin(2n)=a$$

Then

$$\lim_{n\to\infty}\cos n=\sqrt{1-a^2}$$

(I understand it is one case) Then

$$\lim_{n\to\infty}\cos(2n)=\sqrt{1-a^2}$$

Then on one hand

$$a=\lim_{n\to\infty}\sin(2n)=\lim_{n\to\infty}(2\sin n\cos n)=2a\sqrt{1-a^2}$$

$$a\in\left\{\frac{\sqrt{3}}{2},\ -\frac{\sqrt{3}}{2}\right\}$$

and on other hand

$$\sqrt{1-a^2}=\lim_{n\to\infty}\cos(2n)=\lim_{n\to\infty}(\cos^2n-\sin^2n)=1-a^2-a^2=1-2a^2$$

$$1-2a^2>0\ \Leftrightarrow\ a^2<\frac12\ \Leftrightarrow\ a\in\left(-\frac{\sqrt{2}}{2},\ \frac{\sqrt{2}}{2}\right)$$

[Maslov]

gefest_md в сообщении #530915 писал(а):

Cauchy sequence intrigues me. For the rest I can not apply this hint.

$|\sin(n+2) - \sin n| = 2 \sin 1 ~|\cos (n+1)| < \varepsilon \Rightarrow \lim \limits_{n\to\infty} \cos n = 0$

$\cos (n+1) = \cos n \cos 1 - \sin n \sin 1 \Rightarrow $
$\lim\limits_{n\to\infty} \sin n = \dfrac 1 {\sin 1} (\cos 1 \lim\limits_{n\to\infty}\cos n - \lim\limits_{n\to\infty}\cos(n+1)) = 0$

So if $\{\sin n\}$ converges, then both $\{\sin n\}$ and $\{\cos n\}$ converge to $0$, that leads to the contradiction $\lim\limits_{n\to\infty}(\sin^2 n + \cos^2 n) = 0$

[Oleg Zubelevich]

The substance of the problem has been remained out of the sight

[gefest_md]

Oleg Zubelevich в сообщении #531155 писал(а):

The substance of the problem has been remained out of the sight

What do you mean? One might like how I prooved that limit cannot be positive if it is demanded not to pass to limit and not to appeal to the graphic of sine.

[Oleg Zubelevich]

now prove that a set $\{n\pmod{2\pi},\quad n\in\mathbb{N}\}$ is dense in $[0,2\pi]$

[ewert]

Maslov в сообщении #531144 писал(а):

$\cos (n+1) = \cos n \cos 1 - \sin n \sin 1 \Rightarrow $
$\lim\limits_{n\to\infty} \sin n = \dfrac 1 {\sin 1} (\cos 1 \lim\limits_{n\to\infty}\cos n - \lim\limits_{n\to\infty}\cos(n+1)) = 0$

Лучше так:

$|\cos(n+2) - \cos n| = 2 \sin 1 ~|\sin (n+1)|\to0\ \Rightarrow \ \lim \limits_{n\to\infty} \sin n = 0$