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Question Number 155450 by SANOGO last updated on 30/Sep/21

Answered by Kamel last updated on 30/Sep/21

$$L={\underset{n\to +\mathrm{\infty }}{lime}}^{\frac{2}{2n}\underset{k=1}{\sum ^{2n-1}}Ln\left(sin\left(\frac{k\pi }{2n}\right)\right)}=e^{2{\int }_0^{1}Ln\left(sin\left(\pi x\right)\right)dx}=e^{-2Ln\left(2\right)}=\frac{1}{4}$$

Commented by SANOGO last updated on 01/Oct/21

$$mercibien$$

Answered by Ar Brandon last updated on 30/Sep/21

$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }{(\sin \frac{\pi }{2n}\sin \frac{2\pi }{2n}\sin \frac{3\pi }{2n}\cdot \cdot \cdot \sin \frac{2(n-1)\pi }{2n})}^{\frac{1}{n}}$$$$\ln \mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^{2n-1}}\ln \left(\sin \frac{k\pi }{2n}\right)={\int }_0^2\ln \left(\sin \frac{x}{2}\pi \right)dx$$$$=\frac{2}{\pi }{\int }_0^{\pi }\ln \left(\sin u\right)du=\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\ln \left(\sin u\right)du$$$$=-\frac{4}{\pi }\cdot \frac{\pi \ln 2}{2}=-2\ln 2\Rightarrow \mathcal{L}=e^{-2\ln 2}=e^{\ln \frac{1}{4}}=\frac{1}{4}$$

Commented by SANOGO last updated on 01/Oct/21

$$mercibien$$

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