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
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Hint: Multiply the limit by $\pi$. Then you have $n$ strips of width $\pi/n$ on the function $\sin x$ from $0$ to $\pi$. – Parcly Taxel Apr 05 '20 at 13:00
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https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Apr 05 '20 at 13:04
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Or https://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2/469886#469886 – lab bhattacharjee Apr 05 '20 at 13:05
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@ParclyTaxel Thanks ! that did the trick. – mr.pink Apr 05 '20 at 13:22
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Well, you can do it without using integrals :
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 $$ \lim_{n\to +\infty}{\frac{1}{n}\sum_{k=1}^{n}{\sin{\left(\frac{\pi}{n}\right)}}}=\lim_{n\to +\infty}{\frac{1}{n}\cot{\left(\frac{\pi}{2n}\right)}}=\frac{2}{\pi} $$
CHAMSI
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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: $$\lim_{n\rightarrow+\infty}\frac{\pi}{n}\sum_{i=0}^{n}\sin(i\frac{\pi}{n})=\int_{0}^{\pi}\sin(x)dx=2\implies \lim_{n\rightarrow+\infty}\frac{1}{n}\sum_{i=0}^{n}\sin(i\frac{\pi}{n})=\frac{2}{\pi}$$
