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I have asked this question in math.stackexchange without any answer, so I have decided to post it here too.

Recently I was playing around with the sequence $$\frac{1}{n\sin(n)},\ n\in\mathbb{N}.$$

After some computations, I was led to the following question: let $p_n,q_n$ be two sequences of natural numbers such that $$\left|\frac{p_n}{q_n}-\frac{\pi}{2}\right|<\frac{1}{q_n^2}.$$

Can we find a subsequence of $q_n$ composed only by even number? Or is it the case that $q_n$ is odd except for a finite number of indices $n$?

Any idea or reference is welcome.

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    $\begingroup$ This has to do with structure of partial quotients of $\pi/2$, so my gut feeling is that no one on this world knows the answer to your question. $\endgroup$ – Wojowu Mar 29 '16 at 14:31
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The highlighted question is not quite what I expected. If there are infinitely many convergents to $\pi/2$ with odd denominators, then you could make the sequence consist of just these, and the answer is "no," you can't necessarily pass to an infinite subsequence with all even denominators. Every irrational number has infinitely many convergents with odd denominators, since denominators of consecutive convergents are coprime, hence not both even.


A more typical question would be to ask whether there are infinitely many convergents to $\pi/2$ which have even denominators. I would be surprised if this were known. We don't know much about the simple continued fraction expansion of $\pi$ and $\pi/2$, and I don't think we can rule out that all of the coefficients after some point are even, which could mean (with appropriate parities of denominators before this point) that there are only finitely many convergents with even denominators.


Another reasonable question to ask is whether there are infinitely many even $q_n$ so that there are integers $p_n$ so that $|p_n/q_n - \pi/2| \lt 1/q_n^2$. This is a different question than the previous because $p_n/q_n$ does not need to be a convergent, or even to be reduced. There are infinitely many such $q_n$, since for there to be only finitely many even denominators among the convergents, all but finitely many coefficients must be even. If so, then since $\pi/2$ is not quadratic, there must be infinitely many coefficients that are not $2$, hence which are at least $4$. If $a_{n+1} \ge 4$ then $|p_n/q_n - \pi/2| \lt 1/(a_{n+1}q_n^2) \le 1/(4q_n^2)$. So, $|2p_n/(2q_n) - \pi/2| \lt 1/(2q_n)^2$. So, there are infinitely many even integers $q_n$ so that $q_n \pi/2$ is within $1/q_n$ of an integer.

We could prove this another way, using Hurwitz's theorem that there are infinitely many convergents $p/q$ to any irrational $\alpha$ so that $|p/q - \alpha| \le 1/(\sqrt{5}q^2) \lt 1/(2q^2)$. Apply this to $\pi$.

$$\begin{eqnarray}\left|\frac{p}{q} - \pi\right| & \lt & \frac{1}{2q^2} \newline \left|\frac{p}{2q} - \frac{\pi}{2}\right| & \lt& \frac{1}{(2q)^2}\end{eqnarray}$$

So, we get an even denominator $2q$ of a good approximation to $\pi/2$ infinitely often.

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  • $\begingroup$ Excellent. That was the application of Hurwitz that I just couldn't quite see. $\endgroup$ – Aeryk Mar 29 '16 at 20:54
  • $\begingroup$ Thank you, That was just what I needed. As a follow up question, do you think that it is possible to find inifinitely many $q$ and $p$ such that $q$ is a multiple of $n\in \mathbb{R}$ (fixed $n$) and $$\left|\frac{p}{q}-\frac{\pi}{n}\right|<\frac{1}{q^2}?$$ $\endgroup$ – Tomás Mar 30 '16 at 13:27
  • $\begingroup$ In the above comment $n\in \mathbb{N}$. $\endgroup$ – Tomás Mar 30 '16 at 18:04
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    $\begingroup$ I think that for any $n$ there are infinitely many, but that it is an open problem. It would be implied by the conjecture that the coefficients of the simple continued fraction for $\pi$ follow the Gauss-Kuzmin distribution. The real numbers that don't have measure $0$. That it is implied by an open problem doesn't mean it is open or hard but I think it is. $\endgroup$ – Douglas Zare Mar 30 '16 at 18:09
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Yes, such a subsequence exists. [Edit: An infinite number of] The continued fraction convergents, $\frac{p_n}{q_n}$, for $\pi/2$ satisfy $$\left|\frac{\pi}{2}-\frac{p_n}{q_n}\right|<\frac{1}{2q_n^2}<\frac{1}{q_n^2}.$$

[Edit: the rest is incorrect, see comment section.]

If $q_n$ is odd, then $$\left|\frac{\pi}{2}-\left(\frac{p_n}{q_n}+\frac{1}{2q_n^2}\right)\right| \le \left|\frac{\pi}{2}-\frac{p_n}{q_n}\right|+\frac{1}{2q_n^2} < \frac{1}{q_n^2}$$ and $$\frac{p_n}{q_n}+\frac{1}{2q_n^2} = \frac{2q_np_n+1}{2q_n^2}$$ has even denominator.

Note: The $\frac{1}{2q_n^2}$ correction term is somewhat arbitrary. You really only need a fraction less than $\frac{1}{2q_n^2}$ with even denominator.

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    $\begingroup$ When the denominator is odd, you change this by adding $1/(2q_n^2)$. Ok, but that increases the denominator a lot, and the result is no longer a good approximation to $\pi_2$. For example, one good convergent to $\pi/2$ is $11/7$ which has an odd denominator. You change it to $11/7+1/98 = 155/98$ which is no longer a good approximation to $\pi/2$ since the difference is about $1/92$ which is much larger than $1/(2 \times 98^2)$. $\endgroup$ – Douglas Zare Mar 29 '16 at 17:43
  • $\begingroup$ @DouglasZare: Yep, I was incorrect. There are infinitely many $p/q$ with $|\pi/2-p/q|<1/(2q^2)$, but not all convergents are this good. Your second comment points out a more fatal flaw (a failure to work with the new denominator) that invalidates the whole argument. I will edit as appropriate. $\endgroup$ – Aeryk Mar 29 '16 at 20:29

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