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I want to calculate the following limit $$\lim_{n\to\infty} \sin\left(\frac{\pi}{n}\right)\sum_{k=1}^n\frac{1}{2+\cos\left(\frac{k\pi}{n}\right)}$$ and I know I can use Riemann sums to transform the limit to a integral, but don't see how to do this in this particular case.

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  • $\begingroup$ Multiply and divide by $\pi/n$ $\endgroup$
    – RRL
    Jun 4, 2017 at 20:35

1 Answer 1

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Hint:$$\lim _{ n\to \infty } \sin \left( \frac { \pi }{ n } \right) \sum _{ k=1 }^{ n } \frac { 1 }{ 2+\cos \left( \frac { k\pi }{ n } \right) } =\lim _{ n\to \infty } \frac { \sin \left( \frac { \pi }{ n } \right) }{ \frac { \pi }{ n } } \frac { \pi }{ n } \sum _{ k=1 }^{ n } \frac { 1 }{ 2+\cos \left( \frac { k\pi }{ n } \right) } =\\ =\pi \lim _{ n\to \infty } \frac { 1 }{ n } \sum _{ k=1 }^{ n } \frac { 1 }{ 2+\cos \left( \frac { k\pi }{ n } \right) } =\pi \int _{ 0 }^{ 1 }{ \frac { dx }{ 2+\cos { \pi x } } } $$

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    $\begingroup$ Hint? This is the complete solution!(+1) $\endgroup$ Jun 4, 2017 at 20:40
  • $\begingroup$ Shouldn't it be $\pi \int _{ 0 }^{ 1 }{ \frac { dx }{ 2+\cos { \pi x } } }$? $\endgroup$ Jun 4, 2017 at 20:47
  • $\begingroup$ @TomaszTarka,thank you it was a typo $\endgroup$
    – haqnatural
    Jun 4, 2017 at 20:48

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