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Prove that $\displaystyle \lim_{n \rightarrow \infty} \left( \displaystyle \sum_{k=1}^{n}\frac{1}{n}\sin\left(\frac{\pi k }{n}\right) \right)=\displaystyle \frac{2}{\pi} $

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$S_n = \sum_{k=1}^{n} \sin(k \theta) = \csc (\theta/2) \sin(n \theta/2) \sin((n+1) \theta/2)$ To see why the above is true, multiply $\displaystyle \sum_{k=1}^{n} \sin(k \theta)$ by $\sin(\theta/2)$ and write it as a difference of cosines, and telescoping will give you the answer.

Taking $\theta = \dfrac{\pi}n$, we get that $S_n = \sum_{k=1}^{n} \sin \left(\dfrac{k\pi}n \right) = \csc \left(\dfrac{\pi}{2n} \right) \sin \left(\dfrac{n \pi}{2n} \right) \sin \left(\dfrac{(n+1) \pi}{2n} \right) = \dfrac{\sin \left( \dfrac{\pi}{2}+\dfrac{\pi}{2n}\right)}{\sin \left(\dfrac{\pi}{2n} \right) }$ Hence, $a_n = \dfrac{S_n}{n} = \dfrac{\sin \left( \dfrac{\pi}{2}+\dfrac{\pi}{2n}\right)}{n\sin \left(\dfrac{\pi}{2n} \right) }$ Now $\lim_{n \to \infty} a_n = \dfrac{\displaystyle \lim_{n \to \infty} \sin \left( \dfrac{\pi}{2}+\dfrac{\pi}{2n}\right)}{\displaystyle \lim_{n \to \infty} n\sin \left(\dfrac{\pi}{2n} \right) } = \dfrac{\sin(\pi/2)}{\pi/2} = \dfrac2{\pi}$

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For fixed $n$, the sum in question is a right-hand Riemann sum approximating $ \int_0^1 \sin(\pi x) dx. $ Since this function is integrable, the limit exists and equals the integral, which is $2/\pi$.