Does |n2cosn||n^2 \cos n| diverge to +∞+\infty?

I was recently exposed to the problem of deciding whether
lim
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.

Answer

(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.


Attribution
Source : Link , Question Author : Community , Answer Author : Community

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