$\sin (n^2)$ diverges

Question

One can prove that $\sin n$ diverges, using the fact that the natural numbers modulo $2\pi$ is dense.

However, the case for $\sin (n^2)$ looks much more delicate since this is a subsequence of the former one. I strongly believe that this sequence is divergent, but cannot prove it.

In general, can one prove that $\sin (n^a)$ diverges for $a>0$?

Show that for any number (say $\delta$), you can always find $n_1,n_2>\delta$ st $\sin(n_1^2)=0$ and $\sin(n_2^2)=1$ or something similar? And you can apply the same idea for the general case — Shuri2060, Jul 27 '17 at 08:43
Just for clarification: do you consider any real positive $a$ or natural only? — CiaPan, Jul 27 '17 at 09:33
Combining this and this shows ${\sin(n^a)}$ diverges for $a\in(0,1)$. — 19021605, Dec 21 '25 at 15:17
For general $a>0$, see Sungjin Kim's answer here: https://math.stackexchange.com/questions/526857/the-limit-of-sinn-alpha — 19021605, Dec 21 '25 at 16:08

score 14 · Answer 1 · answered Jul 27 '17 at 09:46

14

To generalise Michael's answer:

Theorem. For every non-constant polynomial $f(X)\in\Bbb Q[X]$, the sequence $\sin f(n)$ has infinitely many limit points.

Proof. (By induction).

Let $f(X)\in\Bbb Q[X]$ of degree $d$.

If $d=1$, it is well-known that $\{\,f(n)\bmod 2\pi\mid n\in\Bbb N\,\}$ is dense in $[0,2\pi]$ because $\pi$ is irrational. Then $\{\,\sin f(n)\mid n\in\Bbb N\,\}$ is dense in $[-1,1]$.

If $d>1$, the function $g(X)=f(X+1)-f(X)$ is a polynomial $\in\Bbb Q[X]$ of degree $d-1$. If $\sin f(n)$ has only $N$ limit points then $\cos f(n)$ has at most $2N$ limit points and $$\sin g(n)= \sin f(n+1)\cos f(n)-\sin f(n)\cos f(n+1)$$ has at most $4N^4$ limit points, contradicting the induction hypothesis. $\square$

Remark: I am pretty sure that we actually have "dense in $[-1,1]$" instead of just "has infinitely many limit points".

answered Jul 27 '17 at 09:46

Hagen von Eitzen

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I understand that if $f$ and $g$ have limit points $a$ and $b$ then $ab$ could be a limit point of $fg$ so if they have $N$ and $2N$ limit points this could produce $2N^2$ limit points. Also if $f$ and $g$ have $2N^2$ limit points then $f-g$ could produce $4N^4$ limit points since if $f-g=c$ then there could be $2N^2$ choices for $a$ and $2N^2$ for $b$ such that $b-a=c$. However, I don't understand why a limit point of $fg$ is necessarily generated by limit points of $f$ and $g$. We could have $fg$ close to $c$ without $f$ and $g$ assuming values close to limit points. This also goes for $f-g$ – Evariste Dec 22 '25 at 00:33
In other words, if $c$ isn't of the form $ab$ where $a$ and $b$ are limit points of $f$ and $g$, then why can't $c$ be a limit point of $fg$? – Evariste Dec 22 '25 at 00:38
If $f(u_n)g(u_n) \rightarrow c$ and $f(u_n)$ is bounded then it has a limit point, say $a$ and in this case, $g(u_n)$ has limit point $b=\frac{c}{a}$ if $a$ is nonzero, so $c=ab$ and the claim is proven I guess. If $a=0$ and $g(u_n)$ is bounded, then $c=0$ and in this case $c=ab$ and the claim also holds. Is this correct? In our case both $f$ and $g$ are bounded. I don't know if this holds when $f,g$ are not bounded though, but this wasn't trivial for me. Of course in the non-bounded case we have to allow $c=0 \times \infty$ – Evariste Dec 22 '25 at 01:59
If $f(u_n)$ is unbounded then it has $\pm \infty$ as a "limit point" and $g(u_n)$ has limit point zero so I guess we still have $c=ab$ with $ab = 0 \times \infty$. Therefore the claim $fg$ has limit point $c$ implies $c=ab$ with $a,b$ limit points of $f,g$ holds in full generality (?). Maybe this is trivial but googling it yielded nothing... which is surprising. Maybe this is super obvious or something... – Evariste Dec 22 '25 at 02:10

Empy2 · Accepted Answer · 2017-07-27T11:02:09.663

6

If $\sin n^2$ has a limit $L$, then $\cos n^2$ approaches $\pm \sqrt{1-L^2}$. $$\sin(2n+1)=\sin (n+1)^2\cos n^2-\cos(n+1)^2\sin n^2$$ The right-hand side will have at most four limit points, depending on the signs of the cosines, but the left-hand side is dense in $[-1,1]$

Edit: I think this extends to all $\alpha$.

If $0<\alpha<1$, then $\sin n^\alpha$ is dense because many $n^\alpha$ fit between multiples of $2\pi$.
If $1<\alpha<2$, then $$(n+1)^\alpha-n^\alpha=\alpha n^{\alpha-1}+O(n^{\alpha-2})$$ gets denser as $n$ gets large, so its sine is dense in $[-1,1]$.
If $2<\alpha<3$, then $$(n+2)^\alpha-2(n+1)^\alpha+n^\alpha=\alpha(\alpha-1)n^{\alpha-2}+O(n^{\alpha-3})$$ is dense, and so on.
All the sines of the left-hand sides have limits that can be written as polynomials of $L$ and $\pm\sqrt{1-L^2}$, and so have finitely many values.

edited Jul 27 '17 at 11:02

answered Jul 27 '17 at 09:06

Empy2

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Hi! Could you please elaborate "because many $n^\alpha$ fit between multiples of $2\pi$" (when $0<\alpha<1$)? – Anne Bauval Apr 29 '24 at 22:28
For example $2m\pi\lt n^{1/3}\lt 2(m+1)\pi$ means $(2m\pi)^3\lt n\lt (2(m+1)\pi)^3$, and there are lots of numbers between $(2m\pi)^3$ and $(2(m+1)\pi)^3$ – Empy2 Apr 30 '24 at 03:12
Ok, "many $n^{1/3}$ fit between multiples of $2\pi$", but how does it prove that their $\sin$ are dense? – Anne Bauval Apr 30 '24 at 05:17
$f(x+h)-f(x)=hf'(\xi)$ for some $\xi$ between $x$ and $x+h$. Let $f(x)=x^\alpha$ and $h=1$ then the difference between $f(n)$ and $f(n+1)$ is less than $\alpha n^{\alpha-1}$ which gets smaller as $n$ gets bigger. – Empy2 Apr 30 '24 at 08:22
This is another argument, which still doesn't seem to answer my question. Never mind, thank you for your time and patience. – Anne Bauval Apr 30 '24 at 10:28
"because many $n^{\alpha}$ fit between multiples of $2\pi$"(when $0<\alpha<1$) how do you show this rigorously? – 19021605 Dec 21 '25 at 16:28
Because many $n$ fit between $(2\pi k)^{1/a}$ and (2\pi(k+1))^{1\a} because the derivative of $(2\pi x)^{1/a}$ increases above any number you want. – Empy2 Dec 22 '25 at 01:53

score 2 · Answer 3 · edited Sep 22 '23 at 13:24

2

There is an elementary method. Assume $\sin n^2\to A$, then $\cos 4n^2\to B$ for some $B$ satisfies $A^2+B^2=1$. Now we take limit on both sides of $$ \sin(100n^2)=\sin(36n^2) \cos(64n^2)+\sin(64n^2) \cos(36n^2); $$ $$\cos(100n^2)=\cos(36n^2)\cos(64n^2)-\sin(64n^2)\sin(36n^2)$$ to get $A=2AB,B=B^2-A^2$, hence $A=0,B=1$. But $\sin 4n^2,\sin 4(n+1)^2\to0$ implies $\sin(8n+4)\to0$, hence $$ |\sin 8|\le|\sin(8n+4)|+|\sin(8n-4)|\to0, $$ which is a contradiction. It seems this method cannot be generalized to solve $\sin n^\alpha$.

edited Sep 22 '23 at 13:24

Riemann

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answered Nov 08 '22 at 02:13

P0lyno3ial

176

To add more details in case anyone gets confused (which I did initially): By passing to a subsequence, we may assume $\cos n^2\to B$ for some $A^2+B^2=1$. Then by passing to subsequences again, we have $\cos k^2n^2\to B$ for any $k\in\Bbb Z_{>0}$. Then proceed as in OP's argument up to $\sin(8n+4)=\sin(4(n+1)^2-4n^2)=\sin(4(n+1)^2)\cos(4n^2)-\sin(4n^2)\cos(4(n+1)^2)\to0$. Then $|\sin8|=|\sin((8n+4)-(8n-4))|=|\sin(8n+4)\cos(8n-4)-\sin(8n-4)\cos(8n+4)|\leq|\sin(8n+4)|+|\sin(8n-4)|\to0$. – 19021605 Dec 21 '25 at 16:05

$\sin (n^2)$ diverges

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