How prove this $\lim_{n\to \infty}\sin{n^m}$ divergent.

Solution 1:

Suppose $\lim_{n \to \infty} \sin(n^m) = L \in [-1;1]$.

Let $k \ge 3$ be an odd integer. Since $(\sin((kn)^m))$ is a subsequence of $(\sin(n^m))$, it also converges to $L$.
But we have a relation $\sin(k^m x) = P_k(\sin x)$ where $P_k$ is some polynomial. Hence $(\sin((kn)^m) = P_k(\sin(n^m)) \to P_k(L)$, and so, $P_k(L) = L$. Moreover, for any $n_0$, the sequence $u_n = \sin(k^{nm}n_0)$, which can be recursively defined by $u_0 = \sin(n_0)$ and $u_{n+1} = P_k(u_n)$, has to converge to $L$. So not only is $L$ a fixed point of $P_k$, it probably should be an attractive fixed point too.

First, we look at what can possibly be a fixpoint of all those $P_k$.

We know $L = \sin(l)$ for some $l$ satisfying $\sin(k^ml) = \sin(l)$. But $\sin(k^ml) = \sin(l) \iff k^ml \equiv l \pmod {2\pi} \text{ or } k^ml + l \equiv \pi \pmod {2\pi} \\ \iff l = 2a\pi/(k^m-1) \text{ or } l = (2b+1)\pi/(k^m+1)$.

In any case, $l$ must be a rational multiple of $\pi$ : $l = a\pi/b$ for some coprime integers $a$ and $b$ where $b$ divides $k^{2m}-1$.

Suppose $p$ is an odd prime dividing $b$. Then, picking $k=p$, $p$ has to divide $p^{2m}-1$, which is obviously impossible. So $b$ must be a power of $2$.

We can do a similar thing with $(\sin(2^mx))^2 = P((\sin x)^2)$ where $P$ is some polynomial, to show that $\sin(2^ml) = \pm \sin(l)$, which implies $(2^m\pm 1)l \equiv 0 \pmod \pi$, and in any case, $b$ has to divide $2^{2m}-1$, which is odd.

Hence $b=1$ and so we must have $L = 0$.

Now, differentiating $\sin(k^mx) = P_k(\sin x)$ at $0$, we get $P'_k(0) = k^m > 1$. So $0$ is a repulsive fixpoint for every $P_k$ : there is a small interval around $0$ where $P_3$ pushes away from $0$ : if $\sin(n^m)$ is small but nonzero, then $\sin((3n)^m) = P_3(\sin(n^m))$ will be further from $0$ than $\sin((3n)^m)$, until it is pushed out of that interval. So there is no way that the sequence converges to $0$, unless $\sin(n^m) = 0$ for all large enough $n$. But since $\pi$ is irrational, this doesn't happen.

This is all complicated and all, what with having to consider every $P_k$ to rule out the sequence converging anywhere. If you think about it, when we are at some $\sin(l)$ away from $0$, we found a $k_l$ such that $P_{k_l}(\sin(l)) \neq \sin(l)$ : there actually is some open interval $I_l$ containing $\sin(l)$ on which we can see that $P_{k_l}(I_l) \cap I_l = \emptyset$ . If the sequence ever visits $I_l$, then we know for sure that it will leave it since it will have to visit $P_{k_l}(I_l)$. Similarly, we know there exists a small interval $I_0$ around $0$ such that $P_3$ acts like multiplication by $3^m$ and eventually sends anything nonzero away from $0$ (though the greater $m$ is, the smaller $I_0$ is)

Since $[-1;1]$ is compact, if you consider a concrete $m$, you can use only a finite number of those intervals to obtain a very explicit argument going like this : $[-1;1]$ is the reunion of some explicit intervals $I_0,I_1,I_2, \ldots, I_q$, where $0 \in I_0$, and where for each $i > 0$ there is an odd integer $k_i$ such that $P_{k_i}(I_i) \cap I_i = \emptyset$, which prevents the sequence from converging inside any of those $I_i$. And $P_3$ prevents the sequence from converging in $I_0$ unless it is exactly $0$.