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$$\sum_{i=1}^{n} \frac{\sin(\frac{\pi i}{2n})}{n}$$ Any ideas on how to evaluate this? Basically, I need to find a limit from this sum, when $n \to \infty$ and the only approach I see is to find a formula from this sum.

VIVID
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Prox
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  • $$\sum _{i=1}^n \frac{\sin \left(\frac{\pi i}{2 n}\right)}{n}=\frac{\cot \left(\frac{\pi }{4 n}\right)+1}{2 n}$$ – Raffaele Nov 14 '20 at 11:59

1 Answers1

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By the definition of the definite integral$$\lim_{n\to \infty}\frac 1n\sum_{i=1}^{n} \sin(\frac{\pi i}{2n}) = \int_0^1\sin\frac{\pi x}{2}dx = -\frac 2\pi\cos\frac{\pi x}{2}\bigg|_0^1=\frac{2}{\pi}$$

Also, see this page.

If you really want to find a formula for your sum, see this page. This should help.

VIVID
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  • Your solution just awesome, but I came to this sum exactly from integral, so I need another solution. Thanks in advance. – Prox Nov 14 '20 at 11:21
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    @Prox I didn't type the full solution, but the link provided should help if you don't mind incorporating complex numbers. – VIVID Nov 14 '20 at 11:37