Accumulation points of $ \{x_n \in \mathbb{R}, n \in \mathbb{N} \ \ | \ x_n = n\sin(n) \}$? A younger student asked me:
What are accumulation points of the following set?
$$ \{x_n \in \mathbb{R}, n \in \mathbb{N} \ \ | \ x_n = n\sin(n) \}$$
I really can't answer this question, could anyone help me?
 A: The irrationality measure $\mu(\pi)$ of $\pi$ is not known better than $2\le\mu(\pi)\le7.6063$, it seems. If we knew $\mu(\pi)>2$ then there are infinitely many fractions with $\left|\frac ab-\pi\right|<\frac1{b^\kappa}$ for some $\kappa>2$. Then $|\sin a\|<|a-b\pi|<\frac 1{b^{\kappa-1}}$ and so $|a\sin a|$ becomes arbitrarily small, making $0$ an accumulation point.
Unfortunately, I can only confirm that $\left|\frac ab-\pi\right|<\frac c{b^2}$ for infinitely many fractions, which results in infinitely many $n$ with  $|n\sin n|<c\pi$. So at least some accumulation point exists.
A: Let $E=\{n\sin n:n\in\mathbb{N}\}$. For first, let we prove that a quite big neighbourhood of the origin belongs to $\bar{E}$ (the closure of $E$). By Hurwitz' theorem there are an infinite number of rational numbers $\frac{p_n}{q_n}$ such that:
$$ \left|\pi-\frac{p_n}{q_n}\right|<\frac{1}{\sqrt{5} q_n^2}, $$
so $\left|\pi q_n-p_n\right|<\frac{1}{q_n\sqrt{5}}$ and:
$$\begin{eqnarray*} p_n \sin p_n &=& p_n \left(\sin(\pi q_n)-\cos(\pi q_n)\cdot|\pi q_n-p_n|+O\left(\frac{1}{q_n^2}\right)\right)\\&=& p_n (-1)^{q_n}|\pi q_n-p_n|+O\left(\frac{1}{q_n}\right)\end{eqnarray*}$$
so there are an infinite number of elements of $E$ in the interval $I=\left(-\frac{\pi}{\sqrt{5}},\frac{\pi}{\sqrt{5}}\right)$.
Assuming that an infinite number of $q_n$s is odd (always true) and an infinite number of $q_n$s is even (it looks reasonable), by evaluating $n\sin n$ in $p_n+p_m$ we have that both $(E-E)$ and $(E+E)$ are dense in $E$.
By recalling the lemma for which an infinite set of points of an interval that is closed by difference is dense in the interval, it follows that $E$ is dense in $\mathbb{R}$.
