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Question Number 48068 by maxmathsup by imad last updated on 18/Nov/18

$$let{u}_n={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^n}$$$$findnatureof\mathrm{\Sigma }{u}_{n}and\mathrm{\Sigma }\frac{u_n}{n^2}and\mathrm{\Sigma }\frac{u_n}{n^3}$$

Commented by Abdo msup. last updated on 19/Nov/18

$$changement{t}^n=xgivet=x^{\frac{1}{n}}\Rightarrow$$$$u_n={\int }_0^{\mathrm{\infty }}\frac{1}{1+x}\frac{1}{n}{x}^{\frac{1}{n}-1}dx=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{n}-1}}{1+x}dx$$$$=\frac{1}{n}\frac{\pi }{sin\left(\frac{\pi }{n}\right)}\Rightarrow u_n=\frac{\pi }{nsin\left(\frac{\pi }{n}\right)}$$$$2)wehave{u}_n\sim \frac{\pi }{n.\frac{\pi }{n}}\Rightarrow u_n\to 1(n\to +\mathrm{\infty })so\mathrm{\Sigma }{u}_{n}diverge$$$$because{u}_{n}dontconvergeto0)also$$$$\frac{u_n}{n^2}\sim \frac{1}{n^2}and\mathrm{\Sigma }\frac{1}{n^2}converge\Rightarrow \mathrm{\Sigma }\frac{u_n}{n^2}converge$$$$\frac{u_n}{n^3}\sim \frac{1}{n^3}and?\mathrm{\Sigma }\frac{1}{n^3}converges\Rightarrow \mathrm{\Sigma }\frac{u_n}{n^3}converges.$$

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

$$t^n={tan}^2\theta {nt}^{n-1}dt=2tan\theta {sec}^2\theta d\theta$$$${\int }_0^{\frac{\pi }{2}}\frac{2tan\theta {sec}^2\theta d\theta }{{sec}^2\theta \times n{\left({tan}^2\theta \right)}^{\frac{n-1}{n}}}$$$$=\frac{2}{n}{\int }_0^{\frac{\pi }{2}}{\left(tan\theta \right)}^{1-\frac{2n-2}{n}}d\theta$$$$\frac{2}{n}{\int }_0^{\frac{\pi }{2}}\frac{{\left(sin\theta \right)}^{\frac{2-n}{n}}}{{\left(cos\theta \right)}^{\frac{2-n}{n}}}d\theta$$$$\frac{2}{n}\int {\left(sin\theta \right)}^{\frac{2-n}{2}}\times {\left(cos\theta \right)}^{\frac{n-2}{n}}d\theta$$$$2p-1=\frac{2-n}{2}p=\frac{4-n}{4}=(1-\frac{n}{4})$$$$2q-1=\frac{n-2}{2}2q=\frac{n}{2}q=\frac{n}{4}$$$$nowformula2{\int }_0^{\frac{\pi }{2}}{\left(sin\theta \right)}^{2p-1}{\left(cos\theta \right)}^{2q-1}d\theta$$$$=\frac{\lceil \left(p\right)\lceil \left(q\right)}{\lceil (p+q)}$$$$=\frac{1}{n}\times \frac{\lceil (1-\frac{n}{4})\lceil \left(\frac{n}{4}\right)}{\lceil \left(1\right)}$$$$=\frac{1}{n}\times \frac{\pi }{sin\left(\frac{n\pi }{4}\right)}=\frac{\pi }{n}\times \frac{1}{sin\left(\frac{n\pi }{4}\right)}$$$$so{u}_n=\frac{\pi }{n}\times \frac{1}{sin\left(\frac{n\pi }{4}\right)}$$

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