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.)


1 Answer 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).

  • $\begingroup$ 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. $\endgroup$ Jun 29, 2012 at 6:10
  • $\begingroup$ 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. $\endgroup$ Nov 25, 2015 at 23:46

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