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

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.)

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).

• Interesting -- I unwittingly stumbled upon an unsolved problem. At least now I know the keyword to find out more (or maybe not to spend too much time chasing this question, but I'm going to read the paper). Thanks for your info. Jun 29 '12 at 6:10
• I got the impulse to look at these references again. I still think it is really cool and creative to quantify irrationality by well-approximation like that; I regarded rationality as a qualitative, binary, 'algebraic' property and left it at that, and would have never have thought to do something like this. It's a similar kind of feeling to when you're used to statements being sharply true or false and someone introduces a really rich generalisation with the notion of probabilities between 0 and 1. Nov 25 '15 at 23:46