Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$.
I am looking for nice ways of proving the divergence of the sequence $\left\{x_n\right\}_{n=1}^{\infty}$ 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 $$I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)$$ and one where they lie in $$J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).$$ If ${x_n}$ converges, then all its subsequences must converge to the same limit, 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?
$\endgroup$ 19 Answers
$\begingroup$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.)
$\endgroup$ 8 $\begingroup$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.
$\endgroup$ $\begingroup$This is a simplified version of the proof posted by Sangchul Lee.
If the limit of $\sin{n}$ is $s$, from $$\sin{(n+1)}= \sin{n}\cos{1}-\cos{n}\sin{1}$$ it follows that $\cos{n}$ is also convergent to a limit $c$. Taking limits in the equality above and the similar one for $\cos{(n+1)}$, we get that $$s=s\cdot\cos{1} + c\cdot\sin{1}$$ and $$c=c\cdot\cos{1} - s\cdot\sin{1}.$$ The simplification is that we don't need to solve the system, just add these two equalities after multiplying them by $s$ and $c$, respectively, to obtain $1=\cos{1}$, a contradiction.
$\endgroup$ 3 $\begingroup$Imagine marching around the circumference of the unit circle in steps of arc length $1$. Then $\sin(n)$ is the $y$ coordinate at the $n$th step. Since $\pi\gt1$, the $y$ coordinate will be positive infinitely often and negative infinitely often. Consequently the limit of $\sin(n)$, if it exists, would have to be $0$. But for the same reason ($\pi\gt1$), $\sin(n)\ge\sin({\pi-1\over2})=\sin(\pi-{\pi-1\over2})$ for infinitely many $n$, and hence the limit, if it exists, would have to be at least $\sin({\pi-1\over2})$, which is greater than $0$. These contradictory requirements show that the limit does not exist.
Remark: This proof does not rely on $\pi$ being an irrational number; nor does it use any trig identities beyond the simplest symmetries of the sine function.
$\endgroup$ $\begingroup$This question was posed today (April 16, 2013) as and Douglas Zare gave a nice trig-identity-based proof in comments there.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$To see that the sequence $\left|\sin(n)\right|$ isn't Cauchy it is by the periodicity of $\sin$ enough to show the following: whenever $x_{0}\in[0,\pi-2], x_{1}=x_{0}+1, x_{2}=x_{0}+2$, we have $$\left|\sin(x_{0})-\sin(x_{1})\right|>\eta \text{ or } \left|\sin(x_{1})-\sin(x_{2})\right|>\eta.$$ But this is obvious because either $x_0,x_1\in[0,\pi/2]$ or $x_1,x_2\in[\pi/2,\pi]$, and $\sin$ is strictly monotone in these intervals. For example in the case $x_{0},x_{1}\in[0,\pi/2]$ we have $$\left|\sin(x_{0})-\sin(x_{1})\right|=\sin(x_1)-\sin(x_0)\leq\sin(\pi/2)-\sin(\pi/2-1)=:\eta,$$ where we estimated using the concavity of sin in the interval $[0,\pi/2]$.
$\endgroup$ $\begingroup$A known resultis that if {x_n} and {y_n} are two sequences such that y_n is divergent and x_n\y_n converges to 0 then x_n also diverges. Take x_n=sin (n) and y_n=n then the result follows.
$\endgroup$ 1
