TITLE: Asymptotic behaviour of sum
QUESTION [22 upvotes]: I would like to evaluate the number $c$ given by
$$
c = \lim_{m\to\infty} \frac{1}{\log m}\sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)}
$$
where $\tau = (1+\sqrt{5})/2$. 
My attempt: my guess was this sum would be dominated by the terms for which $n$ is a Fibonacci number. I considered the sum of this sub series using the relation $F_a \tau = F_{a+1}-(-\tau)^{-a}$, the small angle approximation, and that $F_a\tau^{-a} = 1/\sqrt{5}+\mathrm{O(\tau^{-2a})}$. This yielded
$$
\sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)} \approx
\sum_{a=1}^{\log_\tau m \sqrt{5}} \frac{1}{F_a^2 \sin^2(\pi \tau^{-a})} \approx \sum_{a=1}^{\log_\tau m \sqrt{5}} \frac{5}{\pi^2} = \frac{5}{\pi^2} \frac{\log m \sqrt{5}}{\log \tau}
$$
this would imply $c = 5/(\pi^2 \log \tau)$, however this seems about a factor $1.2$ smaller than the correct value, implying I have missed some important terms.
Is it possible to obtain an exact expression? Or at least a better a tighter lower (and upper) bound?

REPLY [2 votes]: Responding to questions in comments about the behavior of 
$$\lim_{K \to \infty} \frac{1}{\log K} \sum_{n=1}^K \frac{1}{n^2 \sin^2 (n \theta)}$$
for other $\theta$: We can transform this to
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \sum_{m=- \infty}^{\infty}  \frac{1}{n^2 (m-n \theta)^2}$$
as before. For arbitrary $\theta$, I doubt we can say anything. But for almost all $\theta$ (in other words, on the complement of a set of measure $0$), we should be able to.
Let $R(x)$ be the closest integer to $x$ ($R$ stands for "round") and let $\langle x \rangle = x-R(x)$. The inner sum is
$$\frac{1}{n^2 \langle n \theta \rangle^2} + \frac{1}{n^2}\sum_{k=1}^{\infty} \left( \frac{1}{(\langle n \theta \rangle +k)^2}+\frac{1}{(\langle n \theta \rangle -k)^2} \right). \quad (4)$$
The sum in (4) is at most $2 \sum_{k=1}^{\infty} \tfrac{1}{(k-1/2)^2}$, which is less than some constant $C$, and then summming $C/n^2$ converges, so this is wiped out by the $\log K$ factor. So what we really care about is
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \frac{1}{n^2 \langle n \theta \rangle^2}. \ (5)$$
I learned from this answer that the number of $n \leq K$ for which $\langle n \theta \rangle \in (x/n, (x+dx)/n)$ is roughly $dx \log K$. So I would predict that $(5)$ behaves like
$$\int_{x=-\infty}^{\infty} \frac{dx}{x^2}.$$
This, of course, diverges, so I would guess the limit does too. Perhaps I'll think about the actual rate of growth later tonight.