Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question
    
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. –  Willie Wong Oct 7 '11 at 15:18
1  
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. –  Greg Martin Oct 7 '11 at 15:56
    
@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. –  Willie Wong Oct 7 '11 at 16:06
add comment

3 Answers

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.

share|improve this answer
    
@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. –  robjohn Oct 7 '11 at 23:56
add comment

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.

share|improve this answer
    
It is worth taking a look at this Math Overflow thread, and the first answer there: mathoverflow.net/questions/24579/convergence-of-a-series –  Eric Naslund Oct 8 '11 at 16:27
1  
Isn't the continued fraction to look at the one for $\frac{\theta}{2\pi}$? –  robjohn Oct 8 '11 at 16:35
    
thanks @robjohn, I fixed it –  Greg Martin Oct 11 '11 at 8:22
add comment

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.

share|improve this answer
    
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. –  Willie Wong Oct 7 '11 at 18:51
    
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. –  Andrew Oct 7 '11 at 19:11
    
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 $\dfrac{2}{\pi}\log(\theta N)$ –  robjohn Oct 7 '11 at 23:49
    
@robjohn well, whats the asymptotic then. –  Andrew Oct 8 '11 at 4:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.