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Question Number 163658 by mathlove last updated on 09/Jan/22

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[n]{\sin \frac{\pi }{2n}\times \sin \frac{2\pi }{2n}\times \sin \frac{3\pi }{2n}....\times \sin \frac{(n-1)\pi }{n}}=?$$

Answered by Ar Brandon last updated on 09/Jan/22

$$\mathcal{A}=\underset{n\to \mathrm{\infty }}{\lim }\sqrt[n]{\sin \frac{\pi }{2n}\times \sin \frac{2\pi }{2n}\times \sin \frac{3\pi }{2n}....\times \sin \frac{(n-1)\pi }{n}}$$$$\ln \mathcal{A}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\ln \left(\underset{k=1}{\prod ^{n-1}}\sin \left(\frac{k\pi }{2n}\right)\right)=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^{n-1}}\ln \left(\sin \left(\frac{k\pi }{2n}\right)\right)$$$$={\int }_0^1\ln \left(\sin \left(\frac{\pi }{2}x\right)\right)dx=\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\ln \left(\sin u\right)du$$$$=\frac{2}{\pi }(-\frac{\pi \ln 2}{2})=-\ln 2\Rightarrow \mathcal{A}=e^{\ln \left(\frac{1}{2}\right)}=\frac{1}{2}$$

Answered by qaz last updated on 09/Jan/22

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\sqrt[\mathrm{n}]{\sin \frac{\pi }{2\mathrm{n}}\cdot \sin \frac{2\pi }{2\mathrm{n}}\cdot ...\cdot \sin \frac{(\mathrm{n}-1)\pi }{\mathrm{n}}}$$$$=\exp \underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{2\mathrm{n}-2}}\mathrm{lnsin}\frac{\mathrm{k}\pi }{2\mathrm{n}}$$$$={\mathrm{e}}^{{\int }_0^2\mathrm{lnsin}\frac{\pi }{2}\mathrm{xdx}}$$$$=\frac{1}{4}$$

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