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What is $$p_*:=\inf\big\{p\in\mathbb R\colon\,\inf_{n\in\mathbb N}n^p\,\inf_{k\in\mathbb N} |2\sqrt{3n}-9\pi/4-k\pi|>0\big\},$$ where $\mathbb N:=\{1,2,\dots\}$?

In other words, I would like to know how small a real $p$ can be so that for some real $c>0$ and all natural $n,k$ we have $$|2\sqrt{3n}-9\pi/4-k\pi|\ge \frac c{n^p}.$$ One may note that $p_*\ge1/2$, since $\sqrt{n+1}-\sqrt n<n^{-1/2}$ and hence $\inf_{k\in\mathbb N} |2\sqrt{3n}-9\pi/4-k\pi|<\sqrt3\,n^{-1/2}$ for infinitely many $n$, so that $\inf_{n\in\mathbb N}n^p\,\inf_{k\in\mathbb N} |2\sqrt{3n}-9\pi/4-k\pi|>0\big\}=0$ if $p<1/2$.

This question is related to this previous one.

As fedja noted in a comment, this problem seems similar to the problem of determining the irrationality measure of $\pi$, and thus may be just too hard.

Actually, for my purposes it would be enough to know how $p_*$ compares with $9/4$: Is $p_*$ greater than, less than, or equal to $9/4$?

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  • $\begingroup$ Something is fishy: as written, the answer is, obviously, $0$ because any positive power of $n$ would be a disaster: there are plenty of $n$ for which $2\sqrt{3n}$ is at constant distance from the given arithmetic progression with difference $\pi$. You obviously meant something else but I'm not sure what. $\endgroup$
    – fedja
    Mar 30, 2022 at 17:03
  • $\begingroup$ @fedja : Thank you for your comment. This is now fixed, hopefully. $\endgroup$
    – Iosif Pinelis
    Mar 30, 2022 at 17:41
  • $\begingroup$ I still believe that the first $\inf$ should be $\sup$ as it was and the second one $\liminf$. $\endgroup$
    – fedja
    Mar 30, 2022 at 18:04
  • $\begingroup$ @fedja : I think the set $\big\{p\in\mathbb R\colon\,\inf_{n\in\mathbb N}n^p\,\inf_{k\in\mathbb N} |2\sqrt{3n}-9\pi/4-k\pi|>0\big\}$ is of the form $(p_0,\infty)$ or $[p_0,\infty)$ for some real $p_0\ge0$ and hence the supremum of this set is $\infty$. As for the $\inf_n$, replacing it by $\liminf_n$ will not make a difference, I think. $\endgroup$
    – Iosif Pinelis
    Mar 30, 2022 at 18:22
  • $\begingroup$ Ah, you also replaced the condition $<\infty$ by $>0$. I haven't noticed that. Then you're right, of course; my apologies. $\endgroup$
    – fedja
    Mar 30, 2022 at 19:36

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