Question Number 53471 by maxmathsup by imad last updated on 22/Jan/19
$$\left.\mathrm{1}\right){find}\:\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{{n}} {tdt}\:\:\:{with}\:{n}\:{integr}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{U}_{{n}} \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 23/Jan/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{n}−\mathrm{2}} \:\left({tan}^{\mathrm{2}} {t}\:+\mathrm{1}−\mathrm{1}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){tan}^{{n}−\mathrm{2}} {tdt}\:−{U}_{{n}−\mathrm{2}} \:\:\:{and}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){tan}^{{n}−\mathrm{2}} {t}\:{dt}\:=\left[{tant}\:{tan}^{{n}−\mathrm{2}} {t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{tant}\:\left({n}−\mathrm{2}\right){tan}^{{n}−\mathrm{3}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\mathrm{1}−\left({n}−\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{{n}−\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\mathrm{1}−\left({n}−\mathrm{2}\right){U}_{{n}−\mathrm{2}} −\left({n}−\mathrm{2}\right){U}_{{n}} \:\Rightarrow \\ $$$${U}_{{n}} =\:\mathrm{1}−\left({n}−\mathrm{2}\right){U}_{{n}−\mathrm{2}} −\left({n}−\mathrm{2}\right){U}_{{n}} −{U}_{{n}−\mathrm{2}} \:\Rightarrow \\ $$$$\left({n}−\mathrm{1}\right){U}_{{n}} =\mathrm{1}−\left({n}−\mathrm{1}\right){U}_{{n}−\mathrm{2}} \:\Rightarrow \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−{U}_{{n}−\mathrm{2}} \:\:\:{with}\:{n}\geqslant\mathrm{2}\:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:−{U}_{\mathrm{2}{n}−\mathrm{2}} \:\:\:\:\:{and}\:{U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:−{U}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$${let}\:{find}\:{U}_{\mathrm{2}{n}} \:\:{we}\:{have} \\ $$$${U}_{\mathrm{2}{n}} \:+{U}_{\mathrm{2}{n}−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left({U}_{\mathrm{2}{k}} \:+{U}_{\mathrm{2}{k}−\mathrm{2}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:\Rightarrow \\ $$$$−\left({U}_{\mathrm{2}} +{U}_{\mathrm{0}} \right)+{U}_{\mathrm{4}} +{U}_{\mathrm{2}} \:−\left({U}_{\mathrm{6}} \:+{U}_{\mathrm{4}} \right)\:+… \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({U}_{\mathrm{2}{n}−\mathrm{2}} \:+{U}_{\mathrm{2}{n}−\mathrm{4}} \right)+\left(−\mathrm{1}\right)^{{n}} \left({U}_{\mathrm{2}{n}} +{U}_{\mathrm{2}{n}−\mathrm{2}} \right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:\Rightarrow\left(−\mathrm{1}\right)^{{n}} \:{U}_{\mathrm{2}{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}} =\left(−\mathrm{1}\right)^{{n}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}} \\ $$
Commented by Abdo msup. last updated on 23/Jan/19
$${let}\:{find}\:{U}_{\mathrm{2}{n}+\mathrm{1}} \:\:\:{we}\:{have}\:{U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:−{U}_{\mathrm{2}{n}−\mathrm{1}} \:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}+\mathrm{1}} \:+{U}_{\mathrm{2}{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left({U}_{\mathrm{2}{k}+\mathrm{1}} +{U}_{\mathrm{2}{k}−\mathrm{1}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}}\:\Rightarrow \\ $$$$−\left({U}_{\mathrm{3}} +{U}_{\mathrm{1}} \right)+{U}_{\mathrm{5}} +{U}_{\mathrm{3}} \:+….+\left(−\mathrm{1}\right)^{{n}} \left({U}_{\mathrm{2}{n}+\mathrm{1}} \:+{U}_{\mathrm{2}{n}−\mathrm{1}} \right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}}\:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:. \\ $$
Commented by Abdo msup. last updated on 23/Jan/19
$${we}\:{see}\:{that}\:\left({U}_{{n}} \right)\:{is}\:{divergent}\:\:{but}\:{we}\:{have} \\ $$$$\left(−\mathrm{1}\right)^{{n}} \:{U}_{\mathrm{2}{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:=_{{k}−\mathrm{1}={p}} \:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}} \\ $$$$=−\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}}\:\rightarrow−\frac{\pi}{\mathrm{4}}\left({n}\rightarrow+\infty\right) \\ $$$${also}\:\left(−\mathrm{1}\right)^{{n}} \:{U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\rightarrow−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\left({n}\rightarrow+\infty\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19
$${I}_{{n}} =\int{tan}^{{n}−\mathrm{2}} {t}×{tan}^{\mathrm{2}} {tdt} \\ $$$$=\int{tan}^{{n}−\mathrm{2}} {t}\left({sec}^{\mathrm{2}} {t}−\mathrm{1}\right){dt} \\ $$$$=\int{tan}^{{n}−\mathrm{2}} {tsec}^{\mathrm{2}} {tdt}−\int{tan}^{{n}−\mathrm{2}} {tdt} \\ $$$$=\int{tan}^{{n}−\mathrm{2}} {t}×{d}\left({tant}\right)−{I}_{{n}−\mathrm{2}} \\ $$$$=\frac{\left({tant}\right)^{{n}−\mathrm{2}+\mathrm{1}} }{\left({n}−\mathrm{2}+\mathrm{1}\right)}−{I}_{{n}−\mathrm{2}} \\ $$$${so}\:{U}_{{n}} =\mid\frac{\left({tant}\right)^{{n}−\mathrm{1}} }{{n}−\mathrm{1}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −{U}_{{n}−\mathrm{2}} \\ $$$$\boldsymbol{{U}}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−{U}_{{n}−\mathrm{2}} \\ $$$$ \\ $$