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Question Number 50426 by Abdo msup. last updated on 16/Dec/18

$$studytheconvergenceof$$$$U_n={\left(\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\left(sinx\right)}^{\frac{1}{n}}\right)}^n$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$${\frac{1}{\pi }\times 2{\int }_0^{\frac{\pi }{2}}sinx)}^{2\times \left(\frac{1+n}{2n}\right)-1}{\left(cosx\right)}^{2\times \frac{1}{2}-1}dx$$$$formula2{\int }_0^{\frac{\pi }{2}}{sin}^{2p-1}{xcos}^{2q-1}xdx$$$$=\frac{\lceil \left(p\right)\lceil q)}{\lceil p+q)}$$$$so\frac{1}{\pi }\times \frac{\lceil \left(\frac{1+n}{2n}\right)\times \lceil \left(\frac{1}{2}\right)}{\lceil (\frac{1+n}{2n}+\frac{1}{2})}$$$$=\frac{1}{\sqrt{\pi }}\times \frac{\lceil (\frac{1}{2n}+\frac{1}{2})}{\lceil (\frac{1}{2n}+1)}$$$$wait...$$$$$$$$$$$$$$$$$$$$$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$Anotherapproach...$$$$1⩾sinx⩾0$$$$1⩾{\left(sinx\right)}^{\frac{1}{n}}⩾0$$$$\frac{2}{\pi }\times {\int }_0^{\frac{\pi }{2}}1\times dx⩾\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\left(sinx\right)}^{\frac{1}{n}}dx⩾\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}0\times dx$$$$2⩾\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\left(sinx\right)}^{\frac{1}{n}}dx⩾0$$$$2^n⩾{\left[\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\left(sinx\right)}^{\frac{1}{n}}dx\right]}^n⩾0$$

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