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Question Number 53471 by maxmathsup by imad last updated on 22/Jan/19

$$1)find{U}_n={\int }_0^{\frac{\pi }{4}}{tan}^{n}tdtwithnintegr.$$$$2)find{lim}_{n\to +\mathrm{\infty }}{U}_n$$$$3)calculate\sum _{n=0}^{\mathrm{\infty }}{U}_n$$$$$$

Commented by Abdo msup. last updated on 23/Jan/19

$$1)wehave{U}_n={\int }_0^{\frac{\pi }{4}}{tan}^{n-2}({tan}^{2}t+1-1)dt$$$$={\int }_0^{\frac{\pi }{4}}(1+{tan}^{2}t){tan}^{n-2}tdt-U_{n-2}andbyparts$$$${\int }_0^{\frac{\pi }{4}}(1+{tan}^{2}t){tan}^{n-2}tdt={\left[tant{tan}^{n-2}t\right]}_0^{\frac{\pi }{4}}$$$$-{\int }_0^{\frac{\pi }{4}}tant(n-2){tan}^{n-3}(1+{tan}^{2}t)dt$$$$=1-(n-2){\int }_0^{\frac{\pi }{4}}{tan}^{n-2}(1+{tan}^{2}t)dt$$$$=1-(n-2)U_{n-2}-(n-2)U_n\Rightarrow$$$$U_n=1-(n-2)U_{n-2}-(n-2)U_n-U_{n-2}\Rightarrow$$$$(n-1)U_n=1-(n-1)U_{n-2}\Rightarrow$$$$U_n=\frac{1}{n-1}-U_{n-2}withn⩾2\Rightarrow$$$$U_{2n}=\frac{1}{2n-1}-U_{2n-2}and{U}_{2n+1}=\frac{1}{2n}-U_{2n-1}$$$$letfind{U}_{2n}wehave$$$$U_{2n}+U_{2n-2}=\frac{1}{2n-1}\Rightarrow$$$$\sum _{k=1}^n{(-1)}^k(U_{2k}+U_{2k-2})=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}\Rightarrow$$$$-(U_2+U_0)+U_4+U_2-(U_6+U_4)+...$$$$+{(-1)}^{n-1}(U_{2n-2}+U_{2n-4})+{(-1)}^n(U_{2n}+U_{2n-2})$$$$=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}\Rightarrow {(-1)}^n{U}_{2n}=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}\Rightarrow$$$$U_{2n}={(-1)}^n\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}$$

Commented by Abdo msup. last updated on 23/Jan/19

$$letfind{U}_{2n+1}wehave{U}_{2n+1}=\frac{1}{2n}-U_{2n-1}\Rightarrow$$$$U_{2n+1}+U_{2n-1}=\frac{1}{2n}\Rightarrow$$$$\sum _{k=1}^n{(-1)}^k(U_{2k+1}+U_{2k-1})=\sum _{k=1}^n\frac{{(-1)}^k}{2k}\Rightarrow$$$$-(U_3+U_1)+U_5+U_3+....+{(-1)}^n(U_{2n+1}+U_{2n-1})$$$$=\sum _{k=1}^n\frac{{(-1)}^k}{2k}\Rightarrow$$$$U_{2n+1}=\frac{{(-1)}^n}{2}\sum _{k=1}^n\frac{{(-1)}^k}{k}.$$

Commented by Abdo msup. last updated on 23/Jan/19

$$weseethat\left(U_n\right)isdivergentbutwehave$$$${(-1)}^n{U}_{2n}=\sum _{k=1}^n\frac{{(-1)}^k}{2k-1}{=}_{k-1=p}\sum _{p=0}^{n-1}\frac{{(-1)}^{p+1}}{2p+1}$$$$=-\sum _{p=0}^{n-1}\frac{{(-1)}^p}{2p+1}\to -\frac{\pi }{4}(n\to +\mathrm{\infty })$$$$also{(-1)}^n{U}_{2n+1}=\frac{1}{2}\sum _{k=1}^n\frac{{(-1)}^k}{k}\to -\frac{ln\left(2\right)}{2}(n\to +\mathrm{\infty })$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

$$I_n=\int {tan}^{n-2}t\times {tan}^{2}tdt$$$$=\int {tan}^{n-2}t({sec}^{2}t-1)dt$$$$=\int {tan}^{n-2}{tsec}^{2}tdt-\int {tan}^{n-2}tdt$$$$=\int {tan}^{n-2}t\times d\left(tant\right)-I_{n-2}$$$$=\frac{{\left(tant\right)}^{n-2+1}}{(n-2+1)}-I_{n-2}$$$$so{U}_n=\mid \frac{{\left(tant\right)}^{n-1}}{n-1}{\mid }_0^{\frac{\pi }{4}}-U_{n-2}$$$${\mathit{U}}_n=\frac{1}{n-1}-U_{n-2}$$$$$$

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