# Compute the limit of $\lim _{n \rightarrow \infty}\left(\frac{\sum_{k=1}^{n} \sin \left(\frac{\pi k}{n}\right)}{n}\right)$

Compute the limit : $$\lim _{n \rightarrow \infty}\left(\frac{\sum_{k=1}^{n} \sin \left(\frac{\pi k}{n}\right)}{n}\right)$$ I know I have to use integral but I dont really know how to. Thanks in advance

• Hint: Multiply the limit by $\pi$. Then you have $n$ strips of width $\pi/n$ on the function $\sin x$ from $0$ to $\pi$. Apr 5 '20 at 13:00
• math.stackexchange.com/questions/17966/… Apr 5 '20 at 13:04
• Apr 5 '20 at 13:05
• @ParclyTaxel Thanks ! that did the trick. Apr 5 '20 at 13:22

Let $$n$$ be a positive integer different from $$0$$ : $$\sum_{k=1}^{n}{\sin{\left(\frac{k\pi}{n}\right)}}=\mathcal{Im}\left(\sum_{k=1}^{n}{\mathrm{e}^{\mathrm{i}\frac{k\pi}{n}}}\right)=\mathcal{Im}\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{n}}\frac{-2}{\mathrm{e}^{\mathrm{i}\frac{\pi}{n}}-1}\right)=\mathcal{Im}\left(\frac{\mathrm{i}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2n}}}{\sin{\left(\frac{\pi}{2n}\right)}}\right)=\cot{\left(\frac{\pi}{2n}\right)}$$
Thus $$\lim_{n\to +\infty}{\frac{1}{n}\sum_{k=1}^{n}{\sin{\left(\frac{\pi}{n}\right)}}}=\lim_{n\to +\infty}{\frac{1}{n}\cot{\left(\frac{\pi}{2n}\right)}}=\frac{2}{\pi}$$
if you want to use the integral then you have to use this formula: $$\int_{a}^{b}f(x)dx=\lim_{n\rightarrow+\infty}\frac{b-a}{n}\sum_{i=0}^{n}f(a+i\frac{b-a}{n})$$ in this case $$a=0 \ ,b=\pi \text{ and } f=\sin$$ we just need to multiply the expression by $$\pi$$ thus: $$\lim_{n\rightarrow+\infty}\frac{\pi}{n}\sum_{i=0}^{n}\sin(i\frac{\pi}{n})=\int_{0}^{\pi}\sin(x)dx=2\implies \lim_{n\rightarrow+\infty}\frac{1}{n}\sum_{i=0}^{n}\sin(i\frac{\pi}{n})=\frac{2}{\pi}$$