Does $\left(n^2 \sin n\right)$ have a convergent subsequence?

Solution 1:

This seems very closely related to the irrationality measure of $\pi$. Fix some exponent $\mu$, and suppose that there are infinitely many $p_n, q_n$ with

$$ \left| \pi - \frac{p_n}{q_n} \right| < \frac{1}{(q_n)^\mu} \, .\,\,\, (*) $$

Then essentially the same argument you gave shows that the subsequence $\left\{x_{p_n}\right\}$ is bounded for any $\alpha \leq \mu-1$.

Conversely, suppose we have a bounded subsequence $\left\{x_{p_n}\right\}$ of $n^\alpha \sin n$, so $(p_n)^\alpha |\sin p_n| < K$ for some fixed $K$ and all $n$. Choose $q_n$ so that $|p_n - q_n \pi| < \frac{\pi}{2}$. Then $$|p_n - q_n \pi| < \frac{\pi}{2} |\sin (p_n-q_n \pi)| < \frac{K \pi}{2(p_n)^{\alpha}} \, ,$$

so $(*)$ holds infinitely often for any $\mu < \alpha + 1$.

According to the Mathworld page I linked to above, all we know about the irrationality measure of $\pi$ is that it's between $2$ and $7.6063$, so your specific problem (which requires that we compare it to $3$) is unsolved.

EDIT: I can't find an online version of the 2008 paper by Salikhov that proves the 7.6063 bound, but here's a pdf of Hata's earlier paper that shows $\mu(\pi) < 8.0161$, and here's a related MathOverflow question (which also has a link to Hata's paper).