Is $\{\sin^n{(n)}:n\in\mathbb{N}\}$ dense in $[-1,1]$?

Question

I know that it's true that the set $\{\sin{(n)}:n\in\mathbb{N}\}$ is dense in $[-1,1]$ but is the set $\{\sin^n{(n)}:n\in\mathbb{N}\}$ also? I would assume it is but I'm unsure of how to prove this because the $n$th power can change the sign of the terms and reduces their absolute value. On a similar note, is it true that $$\limsup_{n\to\infty}\sin^n{(n)}=1$$ $$\liminf_{n\to\infty}\sin^n{(n)}=-1$$ where $n\in\mathbb{N}$? Both of these results would follow if the associated set is dense in $[-1,1]$ but if the initial statement is false then is it possible to seperately prove the limits above?

I have no university level education so I don't think I would be able to provide context such as my own working etc. but I'm quite interested in seeing a proof of the above results as I cannot find them elsewhere.

Answer 1

Yes! The sequence $\{\sin(n)^n\}_{n\in\mathbb{N}}$ is dense on the interval $[-1,1]$. To prove this, I'll use a result proven by this paper, that for any irrational $\omega$, there's infinitely many odd integers $p,q$ for which $|p-q\omega|<1/q$. Using $\omega=\frac{\pi}{2}$, we get two infinite increasing sequences of odd integers $\{p_n\}_{n\in\mathbb{N}}$ and $\{q_n\}_{n\in\mathbb{N}}$ such that $|p_n-q_n\pi/2|<1/q_n$. With some clever analysis, this leads to $|\sin(p_n)|^{p_n} \to 1$ and thus $\limsup_n |\sin(n)|^n=1$, which I proved in this answer to a related question. To prove density, we consider the values $\sin(kp_n)^{kp_n}$ for odd $k$, which makes the approximation $kp_n \approx kq_n\frac{\pi}{2}$ worse in a very controlled way.

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For convenience, let $e_n = p_n-q_n\frac{\pi}{2}$, which must be bounded by $e_n=\mathcal{O}(1/q_n)=\mathcal{O}(1/p_n)$ and thus $e_n\to 0$. Repeating the techniques of the previously linked question, we perform the following analysis for odd integers $k$ obeying $|ke_n| \approx 0$. $$\begin{align} |\sin(kp_n)| &= \left|\cos\left(kp_n - kq_n\frac{\pi}{2}\right)\right| \\ &= \cos(ke_n) \\ &= 1 - \frac{k^2e_n^2}{2} + \mathcal{O}(k^4e_n^4) \\ |\sin(kp_n)|^{kp_n} &= \left(1 - \frac{k^2e_n^2}{2} + \mathcal{O}(k^4e_n^4)\right)^{kp_n} \\ &= \exp\left(kp_n\cdot\left(-\frac{k^2e_n^2}{2} + \mathcal{O}(k^4e_n^4)\right)\right) \\ &= \exp\left(-\frac{k^3 e_n^2 p_n}{2} + \mathcal{O}(k^5 e_n^4 p_n)\right) \\ &\approx \exp\left(-\frac{k^3 e_n^2 p_n}{2}\right) \end{align}$$

Now take any $\alpha\in(0,1)$, and suppose we want to show that $\alpha$ is an accumulation point of $\{|\sin(n)|^n\}_{n\in\mathbb{N}}$. In light of the above, it seems compelling to solve the equation $\alpha = \exp(-k^3 e_n^2 p_n/2)$ for $k$. However, we also want $k$ to be an odd integer, and ideally we can control the value of $k \mod 4$. This can be accomplished with some mild modifications, inspiring the following definition of the sequence $\{k_n\}_{n\in\mathbb{N}}$. $$ k_n := 4\left\lfloor\left(\frac{-\ln(\alpha)}{32e_n^2p_n}\right)^{1/3}\right\rfloor + (q_n\mod 4)$$

Above, the value $q_n\mod 4$ is always either $1$ or $3$, since $q_n$ was odd, hence $k_n$ is also odd. We also notice that, since $e_n^2 p_n$ is bounded above by $\mathcal{O}(1/p_n)$, we get $k_n$ bounded below by $\Omega(p_n^{1/3})$ and thus $k_n\to \infty$. We can also bound $k_ne_n =\mathcal{O}((e_n/p_n)^{1/3})$ so that $k_ne_n \to 0$ as required. This leads to $\alpha$ being an accumulation point, as demonstrated below. $$\begin{align} k_n &\sim 4\left(\frac{-\ln(\alpha)}{32e_n^2p_n}\right)^{1/3} = \left(\frac{-2\ln(\alpha)}{e_n^2p_n}\right)^{1/3} \\ \lim_{n\to\infty} k_n^3e_n^2p_n &= -2\ln(\alpha) \\ \lim_{n\to\infty} k_n^5e_n^4p_n &= \lim_{n\to\infty} \frac{4\ln(\alpha)^2}{k_np_n} = 0 \\ \lim_{n\to\infty} |\sin(kp_n)|^{kp_n} &= \lim_{n\to\infty} \exp\left(-\frac{k^3 e_n^2 p_n}{2} + \mathcal{O}(k^5 e_n^4 p_n)\right) \\ &= \exp(\ln(\alpha) + 0) \\ &= \alpha \end{align}$$

This shows that $\alpha$ is an accumulation point of $\{|\sin(n)|^n\}_{n\in\mathbb{N}}$. Recall, however, that we always have $k_n \equiv q_n \mod 4$ and consequently $q_nk_n \equiv 1 \mod 4$. This implies $\sin(k_np_n)>0$. $$\begin{align} \sin(k_np_n) &= \cos\left(k_np_n - k_nq_n\frac{\pi}{2}\right) \\ &= \cos(k_ne_n) \to 1 \\ |\sin(k_np_n)| &\sim \sin(k_np_n) \\ \lim_{n\to\infty}\sin(k_np_n)^{k_np_n} &= \alpha \end{align}$$

This works for every $\alpha\in (0,1)$, which proves $\{\sin(n)^n\}_{n\in\mathbb{N}}$ is dense on the interval $[0,1]$. To prove density on $[-1,0]$, simply modify the definition of $k_n$ to instead subtract $q_n\mod 4$ instead of adding. This results in $k_n\equiv -q_n\mod 4$ and thus $k_nq_n\equiv -1\mod 4$. The same analysis as above shows that $|\sin(k_np_n)|^{k_np_n}\to \alpha$, but this time we have $\sin(k_np_n)<0$. Since $p_n$ is odd then likewise $\sin(k_np_n)^{k_np_n}<0$, resulting in $-\alpha$ as an accumulation point. $$\begin{align} \sin(k_np_n) &= -\cos\left(k_np_n - k_nq_n\frac{\pi}{2}\right) \\ &= -\cos(k_ne_n) \to -1 \\ |\sin(k_np_n)| &\sim -\sin(k_np_n) \\ \lim_{n\to\infty}\sin(k_np_n)^{k_np_n} &= -\alpha \end{align}$$

This works for all $\alpha\in(0,1)$, which proves $\{\sin(n)^n\}_{n\in\mathbb{N}}$ is dense on $[-1,0]$ and therefore dense on all of $[-1,1]$.

Answer 2

It is very likely that one cannot prove anything rigorous, but here is at least a heuristic showing that the answer to your question is likely to be positive; that is, $\sin^n(n)$ is dense in $(-1,1)$.

We want to show that any fixed subinterval $(\alpha,\beta)\subseteq(-1,1)$ contains a number of the form $\sin^n(n)$. Assume for simplicity that $\alpha>0$. Then $\sin^n(n)\in(\alpha,\beta)$ means that $\sin(n)\in(\alpha^{1/n},\beta^{1/n})$. Using Taylor's approximation, the interval $(\alpha^{1/n},\beta^{1/n})$ has length $C(1+o(1))n^{-1}$, where $C=\log(\beta/\alpha)$. Assuming uniform "distribution" of $\sin(n)$ in $[-1,1]$, the "probability" that this happens is $0.5C(1+o(1))n^{-1}$. The sum of these "probabilities" diverges (as so does the harmonic series $\sum n^{-1}$), and this shows that the event in question is likely to happen infinitely often; that is, our interval $(\alpha,\beta)$ is likely to contain infinitely many numbers $\sin^n(n)$.