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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.

Another User
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sinbadh
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  • I have not done the calculation, so this is pure speculation. One can express the product of two sines and/or cosines as a sum. – André Nicolas Jan 26 '16 at 02:29

1 Answers1

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

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

Ron Gordon
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  • Excelent!! Indeed my answer was $\int_0^{2\pi}x\sin^2xdx$. Is a typing error? – sinbadh Jan 26 '16 at 02:32
  • @sinbadh: no, mistake on my part. Fixed. – Ron Gordon Jan 26 '16 at 02:33
  • Ron. +1 And congratulations on joining the 100-K Club. Did you receive total consciousness yet? ;-)) - Mark – Mark Viola Jan 26 '16 at 02:43
  • @Dr.MV: Thanks! Actually, since I attained 100K over the weekend, I have been sleeping 3 ft above my bed and have been looking for the Keymaster. Oh, and there is no Ron, only Zuul. – Ron Gordon Jan 26 '16 at 03:07
  • @RonGordon. Very sincere congratulations for joining the 100-K Club ! – Claude Leibovici Jan 26 '16 at 03:49
  • @RonGordon, Congratulations on reaching $100$k. Could you please explain the steps assuming I know http://math.stackexchange.com/questions/1279816/find-the-limit-lim-n-rightarrow-infty-sum-k-1n-frac1kn – lab bhattacharjee Jan 26 '16 at 05:04
  • @ClaudeLeibovici: very sincere thank you! – Ron Gordon Jan 26 '16 at 08:51
  • @labbhattacharjee: Thank you. I'm sorry, but I am not sure what you are asking. – Ron Gordon Jan 26 '16 at 08:52
  • @RonGordon, By comparing with $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ I understood $\dfrac kn$ transforms to $x$ but how $$\sin^22\pi x$$ has been arrived and there is no $\dfrac1n$ right? – lab bhattacharjee Jan 26 '16 at 09:40
  • @labbhattacharjee: one of the sines has $2 \pi (k/n)$ as an argument; the other has $2 \pi (k/n) - (\pi/n)$. The $(\pi/n)$ vanishes compared to $2 \pi (k/n)$ as $n \to \infty$. – Ron Gordon Jan 26 '16 at 10:04