Asymptotic behaviour of sum 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?
 A: 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.
A: $\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cI{\mathcal{I}}$Amazingly, your sum can actually be computed in closed form. The answer is
$$c=\frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$
Notation: Let $\tau = \tfrac{1+\sqrt{5}}{2}$ and $\bar{\tau} = \tfrac{1-\sqrt{5}}{2}$, so $\tau+\bar{\tau}=1$ and $\tau \bar{\tau} = -1$. 
Let $R$ be the ring $\ZZ[\tau]$. The ring $R$ is known to be a PID with unit group $\pm \tau^k$. Let $\cI$ be the set of nonzero ideals of $R$. For $m+n \tau \in R$, set $N(m+n \tau) = m^2+mn-n^2 = (m+n \tau) (m+n \bar{\tau})$; for an ideal $I \subseteq R$ set $N(I) = |R/I|$. The relation between these notations is that $|N(m+n \tau)| = N(\langle m+n \tau \rangle)$.

You want to evaluate
$$\lim_{K \to \infty} \frac{1}{\log K} \sum_{n=1}^K \frac{1}{n^2 \sin^2 (\pi n \tau)}.$$
We recall the identity
$$\frac{\pi^2}{\sin^2 (\pi x)} = \sum_{m=-\infty}^{\infty} \frac{1}{(m-x)^2}$$
to rewrite this as
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \sum_{m=- \infty}^{\infty}  \frac{1}{n^2 (m-n \tau)^2}. \quad (1)$$
All the terms are positive, so we may rearrange the sum at will; we group together terms $m-n \tau$ which generate the same ideal $I$ in $R$, giving
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{I \in \cI} \sum_{\begin{matrix} \langle m-n \tau \rangle=I   \\ 1 \leq n \leq K \end{matrix}} \frac{1}{n^2 (m-n \tau)^2} \quad (2).$$
I assume it is legitimate to exchange the limit and the outer sum in (2) (should be easy, but I haven't checked). So we want to consider
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{\begin{matrix} \langle m-n \tau \rangle=I   \\ 1 \leq n \leq K \end{matrix}} \frac{1}{n^2 (m-n \tau)^2}. \quad (3)$$
If $\gamma$ is a generator of $I$, then the list of all generators is the numbers of the form $\pm \tau^k \gamma$. All of these points lie on the hyperbolas $m^2-mn-n^2 = \pm N(I)$, with asymptotes $m = \bar{\tau} n$ and $m = \tau n$. 
As $k \to - \infty$, we approach the asymptote $m=\bar{\tau} n$. Both $n$ and  $m-n \tau$ grow exponentially, so the contribution from those summands is bounded and is wiped out by the $\log K$ term. 
As $k \to \infty$, we approach the $m = \tau n$ asymptote. The number of terms is $\tfrac{\log K + O(1)}{\log \tau}$ and each of those terms is 
$$\frac{1}{n^2 (m-n \tau)^2} = \frac{(m- \bar{\tau} n)^2}{n^2 N(I)^2} =  \frac{(m/n- \bar{\tau})^2}{N(I)^2}.$$
Since we are approaching the asymptote $m/n = \tau$, the numerator approaches $(\tau - \bar{\tau})^2 = 5$. We have a sum of $\tfrac{\log K+O(1)}{\log \tau}$ terms which approach $\tfrac{5}{N(I)^2}$, so $(3)$ is
$$\frac{5}{\pi^2 (\log \tau) N(I)^2}.$$
Plugging into $(2)$, 
$$c=\frac{5}{\pi^2 \log \tau} \sum_{I \in \cI} \frac{1}{N(I)^2} .$$
That last sum is your "about 1.2"; you only computed the contribution from the ideal $\langle 1 \rangle$. (To see the connection, note that $\tau^k = F_{k} \tau + F_{k-1}$.)

I expected this to be the end of the line, but it turns out this sum can actually be evaluated! 
Recall that the $\zeta$ function of $R$ is defined to be 
$$Z(s) := \sum_{I \in \cI} \frac{1}{N(I)^s}.$$
So we want to evaluate $Z(2)$. 
We know $Z(s)$ factors as
$$Z(s) = \zeta(s) L(s)$$
where $\zeta$ is the Riemann $\zeta$ function and
$$L(s) = \sum_{n=1}^{\infty} \frac{\left( \tfrac{5}{n} \right)}{n^s}.$$
We know that $\zeta(2) = \tfrac{\pi^2}{6}$, so we are left to evaluate $L(2)$. 
Using quadratic reciprocity,
$$\left( \frac{5}{n} \right) = \begin{cases} 0 & n \equiv 0 \bmod 5 \\ 1 & n \equiv \pm 1 \bmod 5 \\ -1 & n \equiv \pm 2 \bmod 5 \end{cases}$$
from which we deduce
$$\left( \frac{5}{n} \right) = \frac{2}{\sqrt{5}} \left( \cos \tfrac{2 \pi n}{5} - \cos \tfrac{4 \pi n}{5} \right).$$
So 
$$L(2) = \frac{2}{\sqrt{5}} \left( \sum_{n=1}^{\infty} \frac{\cos \tfrac{2 \pi n}{5} }{n^2} - \sum_{n=1}^{\infty} \frac{\cos \tfrac{4 \pi n}{5} }{n^2} \right).$$
We now recall that, for $0 \leq x \leq 2 \pi$, we have
$$\sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{(\pi-x)^2}{4} - \frac{\pi^2}{12}.$$
Plugging in $x=2 \pi/5$ and $4 \pi/5$ we get
$$L(2) = \frac{4 \pi^2}{25 \sqrt{5}}.$$
Turning quadratic reciprocity symbols into linear combinations of trigonometric functions, and then recognizing the Fourier series that results, is a standard way to evaluate $L$-functions.
Putting it all together, we deduce
$$c=\frac{5}{\pi^2 \log \tau}  \frac{\pi^2}{6} \frac{4 \pi^2}{25 \sqrt{5}} = \frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$
