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?
  • $\begingroup$ 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. $\endgroup$ – Willie Wong Oct 7 '11 at 15:18
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    $\begingroup$ 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. $\endgroup$ – Greg Martin Oct 7 '11 at 15:56
  • $\begingroup$ @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. $\endgroup$ – Willie Wong Oct 7 '11 at 16:06

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.

  • $\begingroup$ @Willie: I just saw that what you are looking for is the behavior in $\theta$. That is a question that I think relies on the behavior of the continued fraction for $\frac{\theta}{2\pi}$. I will puzzle that, too. $\endgroup$ – robjohn Oct 7 '11 at 23:56

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.


Here is an idea, but I didn't check the details. Rewriting $S$ as $$S(N;\theta)= \sum_{n = 1}^{N} \theta\left| \frac{\sin \theta n}{\theta n} \right|$$ gives (at least for small $\theta$) an integral sum for $$ g(N;\theta)=\int_0^{\theta N}\frac{|\sin{x}|}x\,dx, $$ For example, calculations give $S(10^6;10^{-1})/g(10^6;10^{-1})=0.97\ldots$

And for $m$ dividing $N$ $$g(N;2\pi /m)=2\sum_{n=1}^{2N/m-1}(-1)^{n+1}\text{Si}(n \pi )-\text{Si}(2 \pi N /m),$$ where $\text{Si}(x)=\int_0^x\frac{\sin y}y\,dy\;$. The asymptotic for the sine integral function at infinity is known: $$ \text{Si}(x)=\frac{\pi }{2}-\frac{\cos (x)}{x}-\frac{\sin (x)}{x^2}+O(x^{-2}). $$ The idea is to use it, may be with some refining, to get an estimate.

  • $\begingroup$ The problem with this approach is that I don't see how to take the limit $N\to\infty$ for fixed $\theta$. To keep the approximation by Si valid, you need $\theta = O(N^{-1})$ as $N\to\infty$, but this tells me that, roughly speaking $S(\infty,0) = 0$, which we know already. $\endgroup$ – Willie Wong Oct 7 '11 at 18:51
  • $\begingroup$ I don't understand why. It seems that for small enough $\theta$ the integral sum can be estimated from below say as 1/2 of $g$. Also if $\theta$ is incommensurable with $\pi$ then some terms in the integral sum will be greater than corresponding area under the graph of $g$ and some lesser, equidistributed points etc. It seems quite a possibility to me that the limit of $S(N;\theta)/g(N;\theta)$ as $N\to\infty$ is equal to one. $\endgroup$ – Andrew Oct 7 '11 at 19:11
  • $\begingroup$ The asymptotic estimate for $\operatorname{Si}(x)$ you give relies on cancellation from positive and negative values of $\sin(x)$. Without this cancellation, the integral for $g$ diverges like $\dfrac{2}{\pi}\log(\theta N)$ $\endgroup$ – robjohn Oct 7 '11 at 23:49
  • $\begingroup$ @robjohn well, whats the asymptotic then. $\endgroup$ – Andrew Oct 8 '11 at 4:06

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