Answer to Question #250653 in Real Analysis for saduni

Question

Show that lim 𝑛→∞ 1 /𝑛 [sin( 𝜋/ 𝑛 ) + sin( 2𝜋/ 𝑛 ) + … + sin( 𝑛𝜋/ 𝑛 )] = 2/ 𝜋 .

Answer 1

Solution:

$\begin{aligned} &\lim _{n \rightarrow \infty} \frac{1}{n}\left[\sin \frac{\pi}{n}+\sin \frac{2 \pi}{n}+\ldots+\sin \frac{(n-1) \pi}{n}\right] \\ &\quad=\lim _{n \rightarrow \infty}\left\{\frac{1}{n} \sum_{r=1}^{n-1} \sin \left(\frac{r \pi}{n}\right)\right\}= \int_{0}^{1} \sin \pi x d x \\ &\quad=\left[\frac{-\cos \pi x}{\pi}\right]_{0}^{1} \end{aligned}$

$=\dfrac1{\pi}[-\cos \pi+\cos 0] \\=\dfrac1{\pi}[1+1] \\=\dfrac2{\pi}$

Hence, proved.