Are there any results about approximating irrational numbers by fractions in which the numerator and denominator are both odd? [The question arose in looking for whether $\liminf_{n\to\infty}n\cos n = 0$. The corresponding result for $n\sin n$ comes from rational approximation of $\pi$. But the question for $n\cos n$ seems intrinsically harder, and appears to need some additional tool.]

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    $\begingroup$ Do you mean $n|\cos n|$ perhaps? $\endgroup$ – Rahul Jul 24 '17 at 17:28
  • $\begingroup$ Yes! I believe that the lim inf is 0, but I have still not seen a proof . $\endgroup$ – Chris L Jul 26 '17 at 8:20

Both limits have no inf (I mean liminf = - infinity) just as limsup is + infinity

Indeed even for positive integers as $$-1\leq \cos n \leq +1 \longrightarrow -n\leq n\cos n\leq +n$$ and as $n\to +\infty$ the set of limit values is dense in $(-\infty,\;+\infty)$

Same for sin

Hope this helps

A graph for $n\cos n$ from $10^{10}$ to $10^{10}+500$

  • $\begingroup$ I overlooked that! What I actually want to know is if $1/(n\cos n)$ goes to zero. $\endgroup$ – Chris L Jul 24 '17 at 18:40
  • $\begingroup$ @ChrisL Yes it does, but is a completely different sequence. As $\cos n$ stays bounded in $[-1;\;1]$ denominator goes to $\infty$ and the fraction goes to zero $\endgroup$ – Raffaele Jul 24 '17 at 20:51
  • $\begingroup$ But $\cos n$ can get very small. If it is ever as small as $1/n$ then $n\cos n$ will not be going to $\infty$. $\endgroup$ – Chris L Jul 25 '17 at 9:21
  • $\begingroup$ Indeed my answer says that there is no limit since the value of $n\cos n$ bounces between $-\infty$ and $+\infty$. I add a graph to my answer for some values after $n=10^{10}$ $\endgroup$ – Raffaele Jul 25 '17 at 13:25
  • $\begingroup$ @ChrisL: Yes, $\cos n$ can be small, but it is large often enough that $n \cos n$ can get very large. When you are looking for divergence because of "bouncing around", like here, the terms need to be large infinitely often, not all the time. $\endgroup$ – Ross Millikan Jul 25 '17 at 14:18

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