$\sin (n^2)$ 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 $a>0$?

Answer 1

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.

Answer 2

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. $\square$

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

Answer 3

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 $\sin n^\alpha$.