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Question Number 166699 by qaz last updated on 25/Feb/22

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=\mathrm{n}}{\sum ^{2\mathrm{n}}}\sin \frac{\pi }{\mathrm{k}}=?$$

Answered by mathsmine last updated on 25/Feb/22

$$x-\frac{x^3}{6}⩽sin\left(x\right)⩽x......E$$$$\underset{k=n}{\sum ^{2n}}\frac{\pi }{k}=\underset{k=0}{\sum ^n}\frac{\pi }{n+k}=\frac{1}{n}\underset{k=0}{\sum ^n}\frac{\pi }{1+\frac{k}{n}}=S_n$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n=\pi {\int }_0^1\frac{dx}{1+x}=\pi ln\left(2\right)$$$$\underset{k=n}{\sum ^{2n}}{\left({}_{}\frac{\pi }{n+k}\right)}^3=T_n⩽(n+1){\left(\frac{\pi }{2n}\right)}^3=\frac{{\pi }^3}{8}(\frac{1}{n^2}+\frac{1}{n^3})$$$$E\Rightarrow$$$$S_n-\frac{T_n}{6}⩽\underset{k=n}{\sum ^{2n}}sin\left(\frac{\pi }{k}\right)⩽S_n$$$$\underset{n\to \mathrm{\infty }}{\lim }{S}_n-\frac{T_n}{6}⩽\underset{n\to \mathrm{\infty }}{\lim }\underset{n}{\sum ^{2n}}sin\left(\frac{\pi }{k}\right)⩽\underset{n\to \mathrm{\infty }}{\lim }{S}_n$$$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=n}{\sum ^{2n}}sin\left(\frac{\pi }{k}\right)=\pi ln\left(2\right)$$$$$$

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