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Question Number 168355 by Mathspace last updated on 08/Apr/22

$$let{U}_n={\int }_0^1\left(x^n\right)\sqrt{1-x^{2n+1}})dx$$$$1)findaequivalentof{U}_n(n\sim \mathrm{\infty })$$$$2)studythecomvergenceof\mathrm{\Sigma }{U}_n$$

Answered by Mathspace last updated on 09/Apr/22

$$changemrnt{x}^{2n+1}=tgivex=t^{\frac{1}{2n+1}}$$$$U_n={\int }_0^1{t}^{\frac{n}{2n+1}}\sqrt{1-t}\frac{1}{2n+1}{t}^{\frac{1}{2n+1}-1}$$$$=\frac{1}{2n+1}{\int }_0^1{t}^{\frac{n+1}{2n+1}-1}{(1-t)}^{\frac{1}{2}}dt$$$$=\frac{1}{2n+1}{\int }_0^1{t}^{\frac{n+1}{2n+1}-1}{(1-t)}^{\frac{3}{2}-1}dt$$$$=\frac{1}{2n+1}B(\frac{n+1}{2n+1},\frac{3}{2})$$$$=\frac{1}{2n+1}\frac{\mathrm{\Gamma }\left(\frac{n+1}{2n+1}\right).\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(\frac{n+1}{2n+1}+\frac{3}{2})}$$$$\mathrm{\Gamma }\left(\frac{3}{2}\right)=\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2}$$$$\frac{n+1}{2n+1}=\frac{1}{2}\left(\frac{2n+2}{2n+1}\right)=\frac{1}{2}(1+\frac{1}{2n+1})$$$$=\frac{1}{2}+\frac{1}{2(2n+1)}\sim \frac{1}{2}\Rightarrow$$$$\frac{n+1}{2n+1}+\frac{3}{2}\sim \frac{1}{2}+\frac{3}{2}=2\Rightarrow$$$$U_n\sim \frac{\sqrt{\pi }}{2(2n+1)}\times \frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(2\right)}$$$$\Rightarrow U_n\sim \frac{\pi }{2(2n+1)}\sim \frac{\pi }{4n}$$$$2)\mathrm{\Sigma }\frac{\pi }{4n}isdivervente\Rightarrow \mathrm{\Sigma }{U}_{n}is$$$$divergente$$

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