Rate of divergence for the series $\sum |\sin(n\theta) / n|$

Solution 1:

Since $\{n\theta\pmod{2\pi}\}_{n=1}^N$ is evenly distributed in $[0,2\pi]$ when $\theta$ is not a rational multiple of $\pi$ and $N$ is large enough, and the mean of $|\sin(\theta)|$ is $\frac{2}{\pi}$, I would expect that when $\theta$ is not a rational multiple of $\pi$, asymptotically, $$ \sum_{n=1}^N\frac{|\sin(n\theta)|}{n}\sim\frac{2}{\pi}\log(N)\tag{1} $$ The exact way in which $\{n\theta\pmod{2\pi}\}_{n=1}^N$ is evenly distributed in $[0,2\pi]$ is difficult to pin down, so I don't have a good proof of $(1)$ yet, but I will work on it.

Solution 2:

This isn't really an answer, but hopefully it's a way to arrive at an answer. First of all, let $\|x\|$ denote the distance from $x$ to the nearest integer, so that $\|x\| = \min\{x-\lfloor x\rfloor,\lceil x\rceil-x\}$. Then $\|x\|/|\sin x|$ is bounded below and above by the constants $1$ and $\pi$. Therefore to address your question #2, we can change $|(\sin n\theta)/n|$ to $\|n\beta\|/n$ if we want, where $\beta=\theta/2\pi$.

In this formulation, the problem is much more clearly connected to the continued fraction expansion of $\beta$, which produces a sequence of convergents to $\beta$ - rational numbers $r_k$, with larger and larger denominators $q_k$, that approximate $\theta$ extremely well. It is known that the $q_k$ must increase exponentially (at least as fast as the Fibonacci numbers), but they can increase arbitrarily quickly if $\beta$ has ridiculously good rational approximations.

Roughly speaking, when the $n$ in your sum is between $q_k$ and $q_{k+1}$, you should be able to pretend that $\beta$ is equal to $r_{k+1}$ for the purposes of estimating how close $n\beta$ is to the nearest integer. This should allow you to understand the behavior of the sum as $N$ grows. (The motivation for this approach is that if $\beta$ is exactly a rational number, then the summand is periodic and hence the sum is very predictable.)

While existing results on continued fractions are more geared towards $\|x\|$ than $|\sin x|$ as the basic function, the methods should carry over to $|\sin x|$ as well, I hope.