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63

Part of 2005 Today's Calculation Of Integral

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Today's calculation of Integral 63

Source: 1997 Hiroshima University

7/5/2005
For a positive number xx, let f(x)=lim⁡n→∞∑k=1n∣cos⁡(2k+12nx)−cos⁡(2k−12nx)∣f(x)=\lim_{n\to\infty} \sum_{k=1}^n \left|\cos \left(\frac{2k+1}{2n}x\right)-\cos \left(\frac{2k-1}{2n}x\right)\right| Evaluate lim⁡x→∞f(x)x\lim_{x\rightarrow\infty}\frac{f(x)}{x}
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