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Could you please tell me how to calculate the trigonmertic function limit?

$$\lim_{n\to\infty}\sum_{k=1}^{n}\left|\sin\frac{kx}{n}- \sin\frac{(k-1)x}{n} \right| $$

Thank you so much!

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Is $x$ a constant? –  Herng Yi Feb 11 '13 at 8:35

1 Answer 1

Use the fact that

$$\sin\frac{kx}{n}- \sin\frac{(k-1)x}{n} = 2 \sin \frac{x}{2 n} \cos{\left [\left ( k - \frac{1}{2} \right ) \frac{x}{n} \right ]} $$

Note that the argument of the sine is small in the limit, so we have

$$\lim_{n \rightarrow \infty} \sum_{k=1}^{n}\left|\sin\frac{kx}{n}- \sin\frac{(k-1)x}{n} \right| = \frac{|x|}{n} \lim_{n \rightarrow \infty}\sum_{k=1}^{n} \left |\cos{\left [\left ( k - \frac{1}{2} \right ) \frac{|x|}{n} \right ]}\right |$$

This is of the form of a Riemann integral, equal to

$$|x| \int_0^1 dy \: |\cos{x y}|$$

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