In the following we consider the series $ S(N;\theta)= \sum_{n = 1}^{N} \left| \frac{\sin n\theta}{n} \right| $ parametrized by $\theta$. It is well known that this series (taking the limit $N\to\infty$) diverges for any $\theta\in (0,\pi)$, but of course the series converges trivially for $\theta$ being a multiple of $\pi$.
Question: is anything known about the "rate" at which this series diverges?
Note that I am not asking about the rate in $N$: we have that for any $\theta\in (0,\pi)$, $S(N;\theta) \approx \log N$. What I am interested is the implicit constant in the $\approx$ sign, which will depend on $\theta$.
More precisely, observe we have the following trivial estimate $ S(2N;\theta) \leq \sum_{1}^{2N} \frac{1}{n} < 1 + \log 2 + \log N $ On the other hand, we have a also fairly trivial lower bound using the observation that, assuming WLOG $\theta \leq \pi/2$, at most one of $\{ k\theta, (k+1)\theta\}$ can lie within $(-\theta/2, \theta/2)$ when we mod out by $\pi$, $ S(2N;\theta) \geq \frac12 \sin(\theta/2) \sum_1^N \frac1n \geq \frac12 \sin(\theta/2) \log N $ This shows our assertion that $S(N;\theta)\approx_\theta \log N$.
What I am wondering is what can be said about $ f(N;\theta) = \frac{S(N;\theta)}{\log N} $ which is clearly a continuous function of $\theta$.
- The above shows that $\frac12 \sin(\theta/2) \leq \liminf_{N\to\infty} f(N;\theta) \leq \limsup_{N\to\infty} f(N;\theta) \leq 1$. Does the limit in fact exist? Do we know what it is?
- The lower bound above shows that $\liminf f(N;\theta)$, near $\theta = 0$, has linear asymptotics in $\theta$. Is this sharp? (I am guessing it shouldn't be, looking at how wasteful the lower bound estimate is.) Can someone give an improved bound?