A particular Diophantine approximation of $\pi/2$ 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.
 A:  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.
A: 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.
