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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 $\theta$.

  1. 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?
  2. 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?
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    A fairly easy improvement is instead of considering $\{k\theta,(k+1)\theta\}$, we consider $\{k\theta, \ldots, (k+k_0)\theta\}$ where $k_0+1$ is the largest multiple of $\theta \leq \pi/2$ that is $\leq \pi$. This way we improve the $\frac12$ factor to $1 - \lfloor \pi /\theta + 1\rfloor^{-1}$ or something like that. But as $\theta\to 0$ this doesn't change the linear asymptotics.2011-10-07
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    Comment on your notation: you're using $\sim$ to indicate an order of magnitude. To me, $\sim$ is more commonly used to indicate a true asymptotic formula (that is, it presupposes that the limit in your question 1 exists). I would use $\asymp$ for your order-of-magnitude statement.2011-10-07
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    @Greg: you are right, I wasn't paying attention. I meant to use $\approx$ which (at least in the literature I am familiar with) means both $\lesssim$ and its reverse.2011-10-07

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