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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?

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Hi, Sos440 Is it possible to extend this method for a proof of divergence of $sin(n!)$? – M.R. Yegan Oct 1 '14 at 4:43
up vote 24 down vote accepted

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.)

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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. – Sangchul Lee Nov 17 '12 at 2:49
Very nice argument! So this also proves that $(x_n)=(\cos n)$ cannot converge. – Paul Apr 16 '13 at 21:21
Another quick way to see that, if $\lim_{n \to \infty} \sin(n)$ and $\lim_{n \to \infty} \cos(n)$ exist, then both limits are zero is to note that $\sum_{n=0} \sin(n)$ and $\sum_{n=0}^\infty \cos(n)$ have bounded partial sums. Indeed, $\sum_{k=0}^n \sin(k)$ and $\sum_{k=0}^n \cos(k)$ are the real and imaginary parts of $\sum_{k=0}^n e^{ik} = \frac{e^{i(n+1)} - 1}{e^i - 1}$ which is bounded. – Mike F May 26 '14 at 8:34

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.

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This question was posed today (April 16, 2013) as and Douglas Zare gave a nice trig-identity-based proof in comments there.

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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.

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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.

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[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.


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.


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.

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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.

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This is false. (What happens if you take $x_n = 1$ and $y_n = n$ with the same argument?) – mrf May 26 '14 at 7:59

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