Limit of trigonometric series and integrals

Question

I found this problem many years ago, and I still doubt my solution:

Find $$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n}\sin\left(\frac{2\pi k}{n}\right)\sin\left(\frac{2\pi k-\pi}{n}\right)\sin\left(\frac{\pi}{n}\right)$$

I tried solve it by Riemann sums, but the third $\sin$ is not helpful.

Any help or hint?

Thanks in advanced.

Answer 1

The third sine approaches its argument in the limit as $n \to \infty$. In this case, the sum takes the value

$$\begin{align}\int_0^1 dx \, x \, \sin^2{2 \pi x} &= \frac14 - \frac12 \int_0^1 dx \, x \, \cos{4 \pi x}\\ &= \frac14 + \frac1{8 \pi} \int_0^1 dx \, \sin{4 \pi x} \\ &= \frac14 \end{align}$$