Order of magnitude for trigonometric sum Consider the following sum:
$$ S_N = \sum_{ \substack{ k_1 + k_2 + k_3 =N  \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} \sin\left( \frac{k_1\pi}{N} \right) \sin\left( \frac{k_2\pi}{N} \right) \sin\left( \frac{k_3\pi}{N} \right) $$
The summation indexes $k_1, k_2, k_3$ are integers that satisfy the restriction $ k_1 + k_2 + k_3 =N$, and each one of them ranges between $-(N-2)$ and $N$, but they're also different from $0$.
I would be interested to know if there is any way to estimate the order of magnitude of this quantity for arbitrary large $N$ (for example, in terms of big O notation). Any reference to papers or books with similar topics would also be appreciated
 A: If you apply Gerhard's observation that you can bound each term by $C/N^3$ using $|\sin(x)|\leq|x|$ we get the following upper bound:
\begin{equation}
\begin{aligned}
|S_N|
&\leq \sum_{k_1+k_2+k_3=N\\0<|k_i|\leq N}\frac{\pi^3}{N^3}\\
&\leq\frac{\pi^3}{N^3}\sum_{k_1+k_2+k_3=N\\0<|k_i|\leq N}1\\
&\leq\frac{\pi^3}{N^3}2N^2=\frac{2\pi^3}{N}.
\end{aligned}
\end{equation}
$\frac{\sin(x\pi/N)}{x}\geq 0$ for $-N\leq x\leq N$ so this should be a pretty good upper bound, at least asymptotically. 
Update: I had the further idea to try to approximate the sum with an integral. Numerical tests showed that this gave a very good approximation.
\begin{equation}
\begin{aligned}
&\sum_{k_1+k_2+k_3=N\\|k_i|\leq N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{k_3\pi}{N})}{k_1k_2k_3}\\
&\approx \iiint_{k_1+k_2+k_3=N\\|k_i|\leq N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{k_3\pi}{N})}{k_1k_2k_3}\\
&=\iint_{|k_1|,|k_2|\leq N\\|N-k_1-k_2|\leq N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{(N-k_1-k_2)\pi}{N})}{k_1k_2(N-k_1-k_2)}\\
&=\iint_{|k_1|,|k_2|\leq N\\0<k_1+k_2\leq 2N}\frac{\sin(\frac{k_1\pi}{N})\sin(\frac{k_2\pi}{N})\sin(\frac{(N-k_1-k_2)\pi}{N})}{k_1k_2(N-k_1-k_2)}\\
&=\left(\frac{\pi}{N}\right)\iint_{|x_1|,|x_2|\leq \pi\\0<x_1+x_2\leq 2\pi}\frac{\sin(x_1)\sin(x_2)\sin(\pi-x_1-x_2)}{x_1x_2(\pi-x_1-x_2)}\\
\end{aligned}
\end{equation}
on substituting $x_1=\frac{k_1\pi}{N}$ and $x_2=\frac{k_2\pi}{N}$. The integral on the right  is independent of $N$ and can be evaluated as approximately 3.589. Thus we have 
\begin{equation}
\begin{aligned}
S_N\approx \frac{3.589 \pi}{N}=\frac{11.274}{N}.
\end{aligned}
\end{equation}
I have some numerical tests and this is in very close agreement with the sum for large values of N, with a relative error of less than a few % for $N$ bigger than 50 say. 
Here is a plot of the approximation in Mathematika:

A: I believe the problem can be approached from the generating-functions perspective. 
First we notice that at least two of the $k_i$'s are positive. Based on the symmetry, let's consider the sum with $k_1,k_2>0$ and multiply it by 3. This will cover all cases, but the case with all $k_i>0$ will be counted thrice. Hence,
$$S_N = 3S_N^{(2)} - 2S_N^{(1)},$$
where in $S_N^{(2)}$ sums under the assumption $k_1,k_2>0$, while $S_N^{(1)}$ sums under the assumption $k_1,k_2,k_3>0$.

Let me consider $S_N^{(1)}$, which takes the form:
$$S_N^{(1)} = (\sin\alpha)^3\sum_{k_1+k_2+k_3=N\atop k_1,k_2,k_3>0} \frac{1}{k_1k_2k_3} U_{k_1-1}(\cos\alpha)U_{k_2-1}(\cos\alpha)U_{k_3-1}(\cos\alpha),$$
where $U_k()$ are Chebyshev polynomials of second kind and $\alpha:=\frac{\pi}{N}$.
From the generating function for Chebyshev polynomials we have
$$\sum_{m\geq 1} \frac{U_{m-1}(\cos\alpha)}m t^m =\frac{\arctan\frac{t\sin\alpha}{1-t\cos\alpha}}{\sin\alpha}.$$
Hence,
$$S_N^{(1)} = [t^N]\ (\arctan\frac{t\sin\alpha}{1-t\cos\alpha})^3.$$
So, essentially we got the generating function for $S_N^{(1)}$. It should be possible to get the asymptotic by standard means (e.g., see Section 5.4 in generatingfunctionology).

UPDATE. Here is yet another (more straightforward) take on the original problem.
Let $z:=e^{I\frac{\pi}{N}}$ and notice that $\sin(k\alpha) = -\frac{I}{2}(z^k - z^{-k})$, where $I$ is the imaginary unit. Taking into account symmetry in summation indices and that $z^N=-1$, we get
\begin{split}
S_N &= \frac{I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N  \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} (z^{k_1}-z^{-k_1}) (z^{k_2}-z^{-k_2}) (z^{k_3}-z^{-k_3})\\
&=\frac{I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N  \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} \big( (z^{k_1+k_2+k_3}-z^{-(k_1+k_2+k_3)}) - 3(z^{k_1+k_2-k_3}-z^{-(k_1+k_2-k_3)})\big)\\
&=-\frac{3I}{8} \sum_{ \substack{ k_1 + k_2 + k_3 =N  \\ -(N-2) \leq k_1, k_2 , k_3 \leq N \\ k_1, k_2 , k_3 \neq 0 } } \frac{1}{k_1 k_2 k_3} (z^{2k_3}-z^{-2k_3}).
\end{split}
We will need generating functions:
$$\sum_{k=1}^{N} \frac{t^k}{k} = \int_0^t dx \frac{1-x^N}{1-x},$$
$$\sum_{k=-(N-2)}^{-1} \frac{t^k}{k} = \sum_{k=2}^{N-1}\frac{t^{k-N}}{k-N} = \int_0^t dx\frac{1-x^{N-2}}{(1-x)x^{N-1}},$$
and sum of the two:
$$F_N(t) := \sum_{-(N-2)\leq k\leq N\atop k\ne 0} \frac{t^k}{k} = \int_0^t dx \frac{1-x^{N-2}+x^{N-1}-x^{2N-1}}{(1-x)x^{N-1}} $$
Then
$$S_N = -\frac{3I}{8} [t^N]\ F_N(t)^2\big( F_N(z^2t) - F_N(z^{-2}t)\big)$$
