# Prove the divergence of the sequence $(\sin(n))_{n=1}^\infty$.

I am looking for nice ways of proving the divergence of the sequence $(x_n)_{n=1}^\infty$ defined by $$x_n:=\sin(n).$$ One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals $$I_k:=\left(\frac{\pi}{6}+2\pi(k-1),\frac{5\pi}{6}+2\pi(k-1)\right)$$ and one where they lie in $$J_k:=\left(\frac{7\pi}{6}+2\pi(k-1),\frac{11\pi}{6}+2\pi(k-1)\right).$$ If $(x_n)$ converges, then all its subsequences must converge to the same number, but here the first subsequence has all its values in the interval $\left[\frac{1}{2},1\right]$ while the second has its values in $\left[-1,-\frac{1}{2}\right]$. Contradiction.

Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?

-

Equidistribution argument is very elegant, but it becomes a sledgehammer method when it comes to a question of mere convergence.

A simple argument can reveal the divergence of $(\sin n, n \geq 1)$. Let

$$(x_n, y_n) = (\cos n, \sin n)$$

(I changed the notation for the sake of consistency of notation.) Then the application of the addition formula for trigonometric function or the rotation matrix gives

\begin{align*} x_{n+1} &= x_n \cos 1 - y_n \sin 1 \\ y_{n+1} &= x_n \sin 1 + y_n \cos 1. \end{align*}

Now assume $(y_n)$ converges. Then since $\sin 1 \neq 0$, we have

$$x_{n+1} = (y_{n+1} - y_n \cos 1) \cot 1 - y_n \sin 1$$

and hence $(x_n)$ also converges. Now let $(x_n, y_n) \to (\alpha, \beta)$. Then taking limit to the recursive formula we have

\begin{align*} \alpha &= \alpha \cos 1 - \beta \sin 1 \\ \beta &= \alpha \sin 1 + \beta \cos 1. \end{align*}

Solving this system of linear equations give $(\alpha, \beta) = (0, 0)$. On the other hand, since

$$x_n^2 + y_n^2 = 1,$$

we must have

$$\alpha^2 + \beta^2 = 1,$$

a contradiction! Therefore $(y_n)$ cannot converge. ////

Of course, we can say much more on $(y_n)$. For example, we can show that the set of limit points of $(y_n)$ is exactly $[-1, 1]$, and the Cesaro mean of $(y_n)$ is 0 from Weyl's criterion.

(This proof is from the book Problems in Real Analysis Advanced Calculus on the Real Axis.)

-
This is very nice! –  Jason DeVito Nov 17 '12 at 2:42
Very nice indeed! Thanks! –  Spenser Nov 17 '12 at 2:45
Thanks! Anyway I forgot to comment that this is not my original idea, but the idea that I learned from the book Problems in Real Analysis Advanced Calculus on the Real Axis. –  sos440 Nov 17 '12 at 2:49
Very nice argument! So this also proves that $(x_n)=(\cos n)$ cannot converge. –  Paul Apr 16 at 21:21

If $\alpha$ is an irrational number, the numbers $n\alpha$, considered mod $1$, are dense in $[0,1]$. (A stronger result is that they are equidistributed.) Thus the numbers $2n/\pi$ are dense in $[0,1]$, and therefore the integers are dense mod $\pi/2$. It follows that $\sin n$ diverges.

-

This question was posed today (April 16, 2013) as http://mathoverflow.net/questions/127726/integer-multiples-of-a-irrational-dense-in-r-z and Douglas Zare gave a nice trig-identity-based proof in comments there.

-

Essentially, every point in the interval $[-1, 1]$ is a limit point for the sequence $\left\{ {\sin n} \right\}$. Since there is more than one limit point, the sequence diverges.

-

Definition:

[x] is a rounding function if [x] is the biggest integer such that for a real number x, ⌊x⌋

Theorem 1:

Define x_n=[〖10〗^n x]/〖10〗^n for a real number x. Then lim(n→∞)〖x_n=x〗. Proof:

|x_n-x|=|([〖10〗^n x]-〖10〗^n x)/〖10〗^n |≤1/〖10〗^n

Theorem 2:

|[〖10〗^n π]-〖10〗^n π| diverges.

Proof:

Suppose |[〖10〗^n π]-〖10〗^n π| converges to a real number. This implies that there are integers k between 0 and 10 and K such that for n≥K, the nth decimal place of π is k. This is absurd since π is irrational.

Theorem 3:

sin(n) diverges.

Proof:

If sin[〖10〗^n x] has a limit s, then for any ϵ>0 there exist an integer K such that for n≥K, |s-sin[〖10〗^n x]|< ϵ. However, this implies that |[〖10〗^n π]-〖10〗^n π| converges which is proven to be false.

-