A question about the accumulation points of a weird sequence. Recently I came across a strange problem asking me to find the accumulation points of the sequence: $a_n$=$\sqrt[3]{n}\cdot\sin\sqrt{n}$
I really don't know how to write in detail, although I'm sort of sure that the accumulation points should be all $\mathbb{R}$.
I sincerely hope that you can help me solve the problem!!
 A: Fix any $x \in \mathbb{R}$ and $\varepsilon > 0$. WLOG, assume $\varepsilon < |x|/2$. We want to show that there is some $n$ such that $|a_n - x| < \varepsilon$. 


*

*We intend to describe the ''distribution'' of $\sqrt{n}$. The distance between two adjacent points is $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n} + \sqrt{n+1}} < \frac{1}{2\sqrt{n}}.$$
Thus, $$|\sin\sqrt{n+1} -\sin\sqrt{n}| \leq 1\cdot|\sqrt{n+1}-\sqrt{n}| < \frac{1}{2\sqrt{n}}.$$

*Similarly, $$\sqrt[3]{n+1}-\sqrt[3]{n} < \frac{1}{3\sqrt[3]{n}}.$$

*Hence, $$|a_{n+1} - a_n| < \sqrt[3]{n+1} \frac{1}{2\sqrt{n}}. + \frac{1}{2\sqrt[3]{n}} =: \delta_n.$$
Clearly $\delta_n \to 0$ as $n \to \infty$. 

*Let $I_k = \left\{ n \in \mathbb{N}: \left(k-\frac{1}{2}\right)\pi < \sqrt{n} \leq \left(k+\frac{1}{2}\right)\pi \right\}$. Thus in each $I_k$, where $k$ is even, $\sin\sqrt{n}$ is increasing, so $a_n$ is an increasing sequence in $I_k$ where $k$ is even. Note also that as $k\to \infty$, $$\eta^1_k := \sqrt[3]{\min\{I_k\}}-\sqrt[3]{\left(k-\frac{1}{2}\right)^2\pi^2 } \to 0, $$
$$
\eta^2_k := \sqrt{\min\{I_k\}}-\sqrt{\left(k-\frac{1}{2}\right)^2\pi^2 } \to 0,
$$ and similarly $$\eta^3_k := \sqrt[3]{\left(k+\frac{1}{2}\right)^2\pi^2 } - \sqrt[3]{\max\{I_k\}}\to 0,
$$
$$\eta^4_k := \sqrt{\left(k+\frac{1}{2}\right)^2\pi^2 } - \sqrt{\max\{I_k\}}\to 0. 
$$

*Choose an even $k$ large such that $$ a_{\min\{I_k\}} < -|x|, \quad a_{\max\{I_k\}} >|x|$$ and $\delta_n < \varepsilon/2$ for all $n \in I_k$. The claimed result thus follows. 
