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.