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I was recently exposed to the problem of deciding whether $$ \lim_{n \to +\infty} |n \cos n| = +\infty$$ where the limit is taken over the integers. As $|\cos n|$ oscillates throughout the interval $[0,1]$, it seems plausible that every real number ought to be a limit point of this sequence. But to show the limit doesn't exist, it's enough to show the sequence is less than $2$ infinitely often, and this can be done with continued fractions: we can find arbitrarily large $p$ and $q$ with $q$ odd such that $$ \left| \frac{\pi}{2} - \frac{p}{q} \right| < \frac{1}{q^2} $$ which implies $$ \left| \frac{q-1}{2} \pi + \frac{\pi}{2} - p \right| < \frac{1}{q} $$ making $p$ a good approximation of $\pi/2$ modulo $\pi$, and thus $$ \left| p \cos p \right| < \frac{p}{q} < 2 $$

This method fails if we consider instead the sequence $n^{1+\epsilon} \cos n$, since the final inequality is an increasing function of $n$. In fact, a heuristic statistical argument suggests that $$ \lim_{n \to +\infty} |n^{1 + \epsilon} \cos n| = +\infty $$ really does hold: imagine the $n$'s are actually uniformly randomly selected from $[0, \pi]$, there is a roughly $2p/\pi$ chance that $|\cos n| < p$. For any positive constant $B$, the expected number of terms of $n^{1 + \epsilon} |\cos n|$ which are less than $B$ is $$ \sum_{n=1}^{+\infty} \frac{2B}{\pi n^{1 + \epsilon}} $$ which, in particular, is finite.

Of course, this is just heuristic. Does anyone know of a proof or disproof of the conjecture $$ \lim_{n \to +\infty} |n^{1 + \epsilon} \cos n| = +\infty $$ for positive real numbers $\epsilon$, where the limit is taken only over integers?

EDIT: Since there have been a few erroneous answers, I'll point out that observing $|\cos n|$ oscillates from 0 to 1 is not enough to show the limit does not exist. Consider the sequence $g(n)$ defined by

  • $g(0) = 1$
  • $g(n) = \max\{ g(n-1), n / |\cos n| \} + 1$

Then $g(n)$ is an unbounded increasing sequence such that $|g(n) \cos n| > n$ for all $n$, and thus

$$ \lim_{n \to +\infty} |g(n) \cos n| = +\infty $$

Roughly speaking, $g(n)$ diverges to $+\infty$ faster than $n \pmod \pi$ can approximate $\pi / 2$.

The question of this post is whether $n^2$ (or $n^{1 + \epsilon}$) also diverges to $+\infty$ fast enough.

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    Isn't it linked to the rationality measure of $\pi$ ? – Lierre Mar 31 '12 at 16:07
  • @Lierre: Ah! I have almost no familiarity with irrationality measure, so I didn't think of it. I believe that would completely characterize the issue for $|n^{1+\epsilon} \sin n|$, but using $\cos n$ (or more generally $\sin(n + \theta)$, if we were so inclined) adds an extra wrinkle. Wikiwalking at mathworld.wolfram.com takes me to the "Cookson Hills Series" whose convergence is related to my question with $\epsilon = 0.5$, and is an open question. –  Mar 31 '12 at 16:26
  • @Lierre: If you make an answer out of that, I'll accept it. (unless someone has something to say about the extra wrinkle that comes from dealing with $\cos$ instead of $\sin$) –  Mar 31 '12 at 19:40
  • @Hurkil: Sorry, but I not able to write such an answer, I'm not very familiar with irrationnality measure either. – Lierre Mar 31 '12 at 22:05

1 Answers1

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(editor's note: I've moved the original second section to the top, since the original first section answered the wrong question)

Edit: I missed the exponent of $2$.

Obviously $(6)$ implies, $\displaystyle\limsup_{n\to\infty}|n^2\cos(n)|=\infty$. However, evaluating $\displaystyle\liminf_{n\to\infty}|n^2\cos(n)|$ is much more difficult. A difficult result is that the irrationality measure of $\frac\pi2$ is less than $8.01604539$. This means that for all but a finite number of rational approximations, $$ \left|\frac{p}{q}-\frac\pi2\right|\ge\frac1{q^{8.01604539}}\tag{7} $$ To make the estimate in $(3)$ yield an infinite limit, we would need to show that the irrationality measure of $\frac\pi2$ is less than $3$. Wikipedia says that "The exact irrationality measure of $\pi$ is not known, however in $2008$ Salikhov has given the approximation $7.6063$." This would mean that $\displaystyle\lim_{n\to\infty}|n^2\cos(n)|$ does not exist.


As was shown in this answer, we can find an infinite number of continued fraction approximations of $\frac\pi2=\frac{p}{q}$ with $q$ odd and so that $$ \left|p-q\frac\pi2\right|\le\frac1q\tag{1} $$ For approximations as in $(1)$, we have, by Maclaurin Expansion, $$ (-1)^{(q-1)/2}\cos(p)=-\left(p-q\frac\pi2\right)+O\left(p-q\frac\pi2\right)^3\tag{2} $$ Taking $\liminf$, we get $$ \begin{align} \liminf_{p\to\infty}|p\cos(p)| &\le\lim_{p\to\infty}p\cdot\frac{1}{q}\\ &=\frac{\pi}{2}\tag{3} \end{align} $$ In a similar fashion, we can find an infinite number of continued fraction approximations of $\pi=\frac{p}{q}$ so that $$ \left|p-q\pi\right|\le\frac1q\tag{4} $$ For approximations as in $(4)$, we have, by Maclaurin Expansion, $$ (-1)^q\cos(p)\ge1-\frac12\left(p-q\pi\right)^2\tag{5} $$ Taking $\limsup$, we get $$ \begin{align} \limsup_{p\to\infty}|p\cos(p)| &\ge\lim_{p\to\infty}p\left(1-\frac{1}{2q^2}\right)\\ &=\infty\tag{6} \end{align} $$ Thus, $\displaystyle\liminf_{n\to\infty}|n\cos(n)|\le\frac\pi2$ and $\displaystyle\limsup_{n\to\infty}|n\cos(n)|=\infty$. Therefore, the limit does not exist.


robjohn
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  • You missed the square in $n^2 \cos n$. – David E Speyer Apr 03 '12 at 20:23
  • @David: Thanks. I was working on the edit, mainly researching the best known value for the irrationality measure of $\pi$, soon after posting, when I realized that I missed the exponent. – robjohn Apr 03 '12 at 21:26
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    And this cleans up the part I missed about how to apply irrationality measure to the $\cos$ version of the problem; I think we even get $n^e \sin(n + \rho \pi)$ for any rational $\rho$. One could be so bold to ask about $|n^{\mu-1} \cos n|$ ($\mu$ for the irrationality measure of $\pi$), but not me; I'm quite content with this answer. –  Apr 03 '12 at 21:36
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    It sounds like you're assuming that the irrationality measure of $\pi$ actually is close to $7.6063$. But I think Salikhov's paper only proves that as an upper bound. I think it could well be the case that the actual bound is $2+\epsilon$. (I have not read Salikhov's actual paper; I'm going from the MathSciNet review.) – David E Speyer Apr 04 '12 at 10:57
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    @David: I hadn't assumed anything about the irrationality measure of $\pi$ (other than it is less than $8.01604539$). My conclusion was: to get an infinite limit using $(3)$ we would need the irrationality measure of $\frac\pi2$ to be less than $3$. I merely commented that if the excerpt from Wikipedia were true, it would mean that $\displaystyle\lim_{n\to\infty}|n^2\cos(n)|$ does not exist. – robjohn Apr 04 '12 at 11:38
  • A note at the end of [the MathWorld article on irrationality measure](http://mathworld.wolfram.com/IrrationalityMeasure.html) says "Alekseyev (2011) has shown that the question of the convergence of the Flint Hill series is related to the irrationality measure of $\pi$, and in particular, convergence would imply $\mu(\pi)\le2.5$, which is much stronger than the best currently known upper bound." – joriki Oct 25 '12 at 19:13
  • Nice ${}{}{}{}{}$ – Jyrki Lahtonen Feb 01 '18 at 13:07