I'm wrestling with the following:
Question: For what values of $\alpha > 0$ does the sequence $\left(n^\alpha \sin n\right)$ have a convergent subsequence?
(The special case $\alpha = 2$ in the title happened to arise in my work.) In a continuous setting this would be a very simple question since $x^\alpha \sin x$ achieves every value infinitely often for (positive) $x \in \mathbb{R}$, but I feel ill-equipped for this situation -- I have an eye on the Bolzano-Weierstrass theorem and not much else.
I have shown the answer is affirmative for $\alpha \leq 1$. Here is my idea.
Proof when $0 < \alpha \leq 1$: Define $x_n = n^\alpha \sin n$ for all $n \in \mathbb{N}$. We will find a bounded subsequence $(y_n)$ of $(x_n)$ so the Bolzano-Weierstrass theorem applies. Let $n \geq 1$ be arbitrary; then by Dirichlet's approximation theorem there are $p_n, q_n \in \mathbb{N}$ satisfying $$q_n \leq n \,\,\,\,\,\,\,\,\,\, \text{and} \,\,\,\,\,\,\,\,\,\, |q_n \pi - p_n| < \frac{1}{n}.$$ Take $y_n = x_{p_n} = (p_n)^\alpha \sin p_n$ for this index $n$. Then $$\begin{eqnarray}|y_n| &=& (p_n)^\alpha \left|\sin(q_n \pi - p_n)\right|\\ &<& (p_n)^\alpha \left(\frac{1}{n}\right)\\ &<& \left(q_n \pi + \frac{1}{n}\right)^\alpha \left(\frac{1}{n}\right)\\ &\leq& \left(n \pi + \frac{1}{n}\right)^\alpha \left(\frac{1}{n}\right)\\ &\leq& \left(n \pi + \frac{1}{n}\right) \left(\frac{1}{n}\right)\\ &\leq& \pi + 1\end{eqnarray}$$ so the sequence $(y_n)$ has a convergent subsequence $(z_n)$ by the Bolzano-Weierstrass theorem. But $(z_n)$ is in turn a subsequence of $(x_n)$, which proves the claim.
Unfortunately, my strategy breaks down when $\alpha > 1$ (where I replace $\alpha$ by $1$ in the chain of inequalities). So in this case, can a convergent subsequence be found some other way or not? (I have a suspicion as to the answer, but I don't want to bias people based on heuristics.)