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Here we go with a not-so-trivial problem:

Inspired by another problem that I myself asked here, I came with this more general formulation: Let be the sequence $a(n) = |\cos(n)|^{f(n)}$. Then, when does the sequence $|\cos(n)|^{f(n)}$ converges as $n \rightarrow \infty$, for $n \in \mathbb{N}$? And the only thing that we may vary is $f(n)$. The only constrain for $f(n)$ will be that $f(n)$ increases monotonically and is unbounded.

It is easy to see that the numbers that belong to the open interval $(k\pi, (k+1)\pi)$ tend to $0$ as $f(n) \rightarrow \infty$. And on the other side, all natural numbers will always belong to these intervals (if not, that would be equivalent to claim the rationality of $\pi$). Then, all members of the sequence will tend to $0$... All?? Not necessarily, because by Dirichlet's Approximation Theorem, you can always find an integer that is arbitrarily close to a multiple of $\pi$. And the closer you are to a multiple of $\pi$, the closer would be the following and related function to 1:

$$ \lim_{x \rightarrow k\pi} |\cos(x)| = 1 $$

On the other hand, if instead of considering functions with real numbers, we focus on our sequence over the natural numbers, we can see that not all the integers approximate to a multiple of $\pi$ in the same manner. There is an integer sequence that every member of the sequence is closer to a multiple of $\pi$ that the previous member (https://oeis.org/A046947). This integer sequence increases exponentially and let be $b(n)$ that sequence. I've seen computationally, that $b(n) \approx \pi^n$. So...

As I can see this, here there are to different opposing forces acting one against the other to make converge or diverge our sequence. On one side, the convergence speed of the members of the open intervals $(k\pi, (k+1)\pi)$ towards $0$ ($k\in\mathbb{N}$), which we can control through playing with $f(n)$; and on the other side, the convergence speed of the sequence $b(n)$ to a bigger multiple of $\pi$. but this "process" is fixed and we cannot alter it.

It seems that:

$$ f(n) = 2n \implies \nexists \lim_{n \rightarrow \infty} |\cos(n)|^{f(n)} $$ This problem seem to be solved here in my previous MSE question. This shows an example of $f(n)$ that makes this sequence diverge, but I strongly think that other expressions for $f(n)$ can make this sequence converge to $0$. My intuition tells me that if $f(n)$ is linear, the sequence will diverge always, but I have that feeling that tells me that if $f(n)$ is exponential, like $f(n) = a^n$, then the sequence will converge if $a > a_0$, being $a_0$ some constant (maybe $\pi$?).

So the final question to answer would be:

Let $f(n)$ be a monotonically increasing and unbounded sequence whose expression is known. How must be $f(n)$ so: $$ \lim_{n \rightarrow \infty} |\cos(n)|^{f(n)} = 0? $$

Many thanks in advance!

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  • $\begingroup$ I maybe missing something here, but doesn't irrationality of $\pi$ imply that $|cos(n)|<1$ for any natural $n$? Thus $\lim_{n \rightarrow \infty} |\cos(n)|^n = 0$ $\endgroup$ – Vasya Aug 29 '18 at 14:12
  • $\begingroup$ @Vasya Your conclusion doesn't hold in general. Consider $(\frac{1}{2^{1/n}})^n$ as a counterexample. $\endgroup$ – Jakobian Aug 29 '18 at 14:25
  • $\begingroup$ For example math.stackexchange.com/questions/2871833/… $\endgroup$ – rtybase Aug 29 '18 at 17:09
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An example faster than $2n$

Here, we take $f(n)=n^2$. We claim that there are infinitely many pairs of positive integers $(n,k)$ such that $$ |n-2\pi k|\leq \frac1{k}. $$ This is achieved if $\frac nk$ is a simple continued fraction convergents of $2\pi$.

Then we have $$ \cos\frac1k = 1-\frac1{2k^2} +O(\frac1{k^4}). $$

Thus, the same arguments (in Jack D'Aurizio's answer) as in the link you provided, we see that for any $\epsilon>0$ there are infinitely many pairs $(n,k)$ such that $$ |\cos n|^{k^2}\geq e^{-1/2}-\epsilon. $$ Since $k\asymp n$, we see that $$ |\cos n|^{n^2} $$ is greater than a fixed positive number when $n\rightarrow \infty$ along the numerators of convergents of $2\pi$.

Along with the existence of infinitely many $n$ such that $|\cos n|\leq 1/2$, we see that $$ |\cos n|^{n^2} $$ does not tend to a limit as $n\rightarrow \infty$.

A polynomial function with convergence to zero

Since $\pi$ has a finite irrationality measure (see here), the inequality $$ |n-\pi k| > \frac1{k^{\mu + \epsilon -1}} $$ for all sufficiently large integers $n,k$. Here, we use $\mu=7.6063$.

Let $f(n)=n^{2(\mu+2\epsilon-1)}$. Then there is a fixed positive $c$ such that $$ |\cos n|^{f(n)} \leq \left|1-\frac c{n^{2(\mu+\epsilon -1)}} \right|^{f(n)}\rightarrow 0. $$ as $n\rightarrow\infty$.

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I'm far from being an expert here, but I suspect that there is no explicit answer known. The problem is that if $n$ is very close to $k\pi$ then $|\cos(n)|$ is very close to $1$, so $f(n)$ has to be very large in order to make $|\cos(n)|^{f(n)}$ small.

So given a sequence $f(n)$, in order to determine whether $|\cos(n)|^{f(n)}$ tends to zero we need fairly precise information about the behavior of the sequence $\delta_n$ defined by $\delta_n=\min_{k\in\Bbb Z}|n-k\pi|$; my impression is that we don't know enough about this sequence.

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  • $\begingroup$ I suspect that it can be easy to find a sequence $\gamma_n \geq \delta_n$, but well defined. I suspect that it shouldn't be difficult to find such sequence, and use is to bound or characterize $f(n)$... I think $\endgroup$ – Carlos Toscano-Ochoa Aug 29 '18 at 21:21
  • $\begingroup$ @CarlosToscano-Ochoa You may be right - let us know... $\endgroup$ – David C. Ullrich Aug 29 '18 at 23:54

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