Compute $\lim_{n\to\infty}\frac{1}{n^3}\sum_{1\le i<j<k\le n} \sin\left(\frac{i+j+k}{n}\right)$

Compute

$$\lim_{n\to\infty}\frac{1}{n^3}\sum_{1\le i<j<k\le n} \sin\left(\frac{i+j+k}{n}\right)$$

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\begin{align} \lim_{n \to \infty} \dfrac1{n^3} \sum_{1\le i<j<k\le n} \sin\left(\frac{i+j+k}{n}\right) & = \lim_{n \to \infty} \dfrac16 \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \sin \left(\dfrac{i}n + \dfrac{j}n + \dfrac{k}n\right) \dfrac1{n^3} \end{align} Because each is a Riemann Sum, we can make this into an integral, when we take the limit. The above is true since the region $0 \leq x \leq y \leq z$ divides the unit cube into $1/6$ of the total region. The integral being symmetric should be the same in all six regions. There is a overlap on the boundaries but this is a measure zero set and hence won't affect the integral. \begin{align} \lim_{n \to \infty} \dfrac16 \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \sin \left(\dfrac{i}n + \dfrac{j}n + \dfrac{k}n\right) \dfrac1{n^3} & = \dfrac16 \int_{[0,1]^3} \sin(x+y+z) dx dy dz\\ & = \dfrac43 \sin^4(1/2) (1+2 \cos(1)) \end{align} We could have also directly converted this directly into the integral shown below without the $1/6$ factor i.e. $$\lim_{n \to \infty} \dfrac1{n^3} \sum_{1\le i<j<k\le n} \sin\left(\frac{i+j+k}{n}\right) = \int_{z=0}^{z=1} \int_{y=0}^{y=z} \int_{x=0}^{x=y} \sin(x+y+z) dx dy dz$$ If you add in a factor of $\pi$ in your original expression i.e. $$\dfrac1{n^3} \sum_{1\le i<j<k\le n} \sin\left(\pi \left(\frac{i+j+k}{n} \right)\right) = - \dfrac{4}{3 \pi^3}$$
In case, the factor $1/6$ is confusing, we could have also directly converted this directly into the integral shown below without the $1/6$ factor as I have shown now. –  user17762 Jan 20 '13 at 22:32
You mean in $d$ dimensions? i.e. $\dfrac1{n^d} \sin((i_1+i_2 + \cdots i_d)/n)$ –  user17762 Jan 20 '13 at 22:37
Yes. I mean a generalization in $d$ dimensions. –  Chris's sis the artist Jan 20 '13 at 22:38