Se observă că aceasta este o sumă Riemann.
Care se calculeaza astfel:
[tex]\lim\limits_{n\to \infty} \dfrac{\sum\limits_{k=1}^n \sin \dfrac{k\pi}{2n}}{n} = \lim\limits_{n\to \infty} \dfrac{1}{n}\sum\limits_{k=1}^n \sin \Bigg[\Big(\dfrac{k}{n}\Big)\cdot \dfrac{\pi}{2}\Bigg] = \\ \\ = \displaystyle \int_{0}^1\sin\Big(x\cdot \dfrac{\pi}{2}\Big) = \dfrac{2}{\pi}\int_{0}^1\Big(\dfrac{x\pi}{2}\Big)'\cdot \sin\Big(\dfrac{x\pi}{2}\Big) = \\ \\ = -\dfrac{2}{\pi}\cdot \cos\Big(\dfrac{x\pi}{2}\Big)\Bigg|_{0}^1 = 0 + \dfrac{2}{\pi}\cdot 1 = \boxed{\dfrac{2}{\pi}}[/tex]
