$\sin (n^2)$ diverges

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".

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.