How to show that $\lim_{n \to \infty} \sin(n^2)$ does not exist?

Question

I know that for $n \to \infty, \space \sin(n)$ does not have limit. Is this also true for the sequence $\sin(n^2)$?

Answer 1

Let we give the tough approach. We want to show that the sequence given by $\sin(n^2)$ is dense in the interval $[-1,1]$. For such a purpose, it is enough to show that the sequence given by $e^{in^2}$ is dense in the unit circle, since the projection on the $y$-axis preserves density. But density (truth to be told, equidistribution, that is way stronger) is a consequence of Weyl's inequality/Van Der Corput's trick with $f(x)=x^2$.

Now a variation on the easy, tricky proof. Assuming that $\sin(n^2)\to L$, we must have $$ \lim_{n\to +\infty}\left(\sin(n^2+2n+1)-\sin(n^2-2n+1)\right)=0 $$ or $$ \lim_{n\to +\infty}2\cos(n^2+1)\sin(2n)=0$$ or $$\lim_{n\to +\infty}\left(\cos(n^2)\cos(1)\sin(2n)-\sin(n^2)\sin(1)\sin(2n)\right) = 0 $$ so the sequence given by $b_n=\sin(2n)$ has at most two accumulation points, that is clearly contradicted by the density (or equidistribution) of $e^{2ni}$ in $S^1$.

Answer 2

There is a simple way to go about proving that the limit does not exist, without the much stronger statement that ${n^2}$ is dense mod $2\pi$.

Note that $(n+1)^2-n^2=2n+1$. In other words, the difference between two consecutive squares is dense mod $2\pi$. Note also that $n^2-(n-1)^2=2n-1$.

Let $n^2=\alpha$ and $(2n+1)=\theta$. Our original sequence was $a_n =\sin(n^2)$. Three consecutive terms are $\sin(\alpha-\theta+2)$, $\sin(\alpha)$, and $\sin(\alpha+\theta)$. It is not hard to show that the largest absolute difference between two of them is at least some $r>0$.

Answer 3

If $\sin(n^{2}) \to L$ then $|L| \leq 1$ and let $a = \sin^{-1}L$. Since $\sin(n^{2}) \to L$ as $n \to \infty$ it follows that $$n^{2} = a_{n}\pi + (-1)^{a_{n}}a + c_{n}$$ where $c_{n} \to 0$ as $n \to \infty$ and $a_{n}$ is sequence which takes integer values only. Let's put $b_{n} = (-1)^{a_{n}}a$ and then we can rewrite the above equation as $$n^{2} = a_{n}\pi + b_{n} + c_{n}\tag{1}$$ and note that $a_{n}$ is integer and $b_{n}$ takes only a finite number of values (namely $a$ and $-a$). From $(1)$ we get $$2n + 1 = (a_{n + 1} - a_{n})\pi + (b_{n + 1} - b_{n}) + (c_{n + 1} - c_{n})$$ and we get an equation similar to $(1)$ like $$2n + 1 = d_{n}\pi + e_{n} + f_{n}\tag{2}$$ where $d_{n}$ is integer, $e_{n}$ takes a finite number of values and $f_{n} \to 0$. Next we have $$2 = (d_{n + 1} - d_{n})\pi + e_{n + 1} - e_{n} + f_{n + 1} - f_{n}\tag{3}$$ Now this is possible only when $f_{n + 1} - f_{n} = 0$ for all $n \geq m$ for some specific positive integer $m$. And then we get $$e_{n} = e_{m} + k_{n}\pi + 2n - 2m$$ for all $n \geq m$ where $k_{n}$ is integer. Since $e_{n}$ takes a finite number of values this is possible only when $k_{n}\pi + 2n$ takes a finite number of values. And this is not possible since $\pi$ is irrational. Here we have used the fact that if $\alpha$ is irrational then the set $A = \{n\alpha - [n\alpha]\mid n \in \mathbb{Z}\}$ is dense in $[0, 1]$.

Thus $\sin(n^{2})$ can't tend to any limit $L$. The technique used in the above proof is very general and taken from Hardy's A Course of Pure Mathematics who attributes it to Albert E. Ingham. Hardy thus proves that if $P(n)$ is a polynomial of positive degree with integer coefficients and $\sin(P(n)\theta\pi) \to 0$ then $\theta$ must be rational.

There is another problem in Hardy's book of similar variety (but with significantly more complexity) which deals with the behavior of $\sin(a^{n}\theta\pi)$ where $a$ is an integer greater than $1$.

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Update: After I wrote this post I found a past comment (to OP's post) by user1952009 which essentially uses the same argument. So this answer can be taken as an exposition of the comment from user1952009.