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

1)find U_n = ∫_0 ^(π/4) tan^n tdt with n integr . 2) find lim_(n→+∞) U_n 3) calculate Σ_(n=0) ^∞ U_n

$$\left.\mathrm{1}\right){find}\:\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{{n}} {tdt}\:\:\:{with}\:{n}\:{integr}\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{U}_{{n}} \\ $$$$ \\ $$

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

1) we have U_n =∫_0 ^(π/4) tan^(n−2) (tan^2 t +1−1)dt =∫_0 ^(π/4) (1+tan^2 t)tan^(n−2) tdt −U_(n−2) and by parts ∫_0 ^(π/4) (1+tan^2 t)tan^(n−2) t dt =[tant tan^(n−2) t]_0 ^(π/4) −∫_0 ^(π/4) tant (n−2)tan^(n−3) (1+tan^2 t)dt =1−(n−2)∫_0 ^(π/4) tan^(n−2) (1+tan^2 t)dt =1−(n−2)U_(n−2) −(n−2)U_n ⇒ U_n = 1−(n−2)U_(n−2) −(n−2)U_n −U_(n−2) ⇒ (n−1)U_n =1−(n−1)U_(n−2) ⇒ U_n =(1/(n−1)) −U_(n−2) with n≥2 ⇒ U_(2n) =(1/(2n−1)) −U_(2n−2) and U_(2n+1) =(1/(2n)) −U_(2n−1) let find U_(2n) we have U_(2n) +U_(2n−2) =(1/(2n−1)) ⇒ Σ_(k=1) ^n (−1)^k (U_(2k) +U_(2k−2) )=Σ_(k=1) ^n (((−1)^k )/(2k−1)) ⇒ −(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) ) =Σ_(k=1) ^n (((−1)^k )/(2k−1)) ⇒(−1)^n U_(2n) =Σ_(k=1) ^n (((−1)^k )/(2k−1)) ⇒ U_(2n) =(−1)^n Σ_(k=1) ^n (((−1)^k )/(2k−1))

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{{n}−\mathrm{2}} \:\left({tan}^{\mathrm{2}} {t}\:+\mathrm{1}−\mathrm{1}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){tan}^{{n}−\mathrm{2}} {tdt}\:−{U}_{{n}−\mathrm{2}} \:\:\:{and}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){tan}^{{n}−\mathrm{2}} {t}\:{dt}\:=\left[{tant}\:{tan}^{{n}−\mathrm{2}} {t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{tant}\:\left({n}−\mathrm{2}\right){tan}^{{n}−\mathrm{3}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\mathrm{1}−\left({n}−\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tan}^{{n}−\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\mathrm{1}−\left({n}−\mathrm{2}\right){U}_{{n}−\mathrm{2}} −\left({n}−\mathrm{2}\right){U}_{{n}} \:\Rightarrow \\ $$$${U}_{{n}} =\:\mathrm{1}−\left({n}−\mathrm{2}\right){U}_{{n}−\mathrm{2}} −\left({n}−\mathrm{2}\right){U}_{{n}} −{U}_{{n}−\mathrm{2}} \:\Rightarrow \\ $$$$\left({n}−\mathrm{1}\right){U}_{{n}} =\mathrm{1}−\left({n}−\mathrm{1}\right){U}_{{n}−\mathrm{2}} \:\Rightarrow \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−{U}_{{n}−\mathrm{2}} \:\:\:{with}\:{n}\geqslant\mathrm{2}\:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:−{U}_{\mathrm{2}{n}−\mathrm{2}} \:\:\:\:\:{and}\:{U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:−{U}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$${let}\:{find}\:{U}_{\mathrm{2}{n}} \:\:{we}\:{have} \\ $$$${U}_{\mathrm{2}{n}} \:+{U}_{\mathrm{2}{n}−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left({U}_{\mathrm{2}{k}} \:+{U}_{\mathrm{2}{k}−\mathrm{2}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:\Rightarrow \\ $$$$−\left({U}_{\mathrm{2}} +{U}_{\mathrm{0}} \right)+{U}_{\mathrm{4}} +{U}_{\mathrm{2}} \:−\left({U}_{\mathrm{6}} \:+{U}_{\mathrm{4}} \right)\:+... \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({U}_{\mathrm{2}{n}−\mathrm{2}} \:+{U}_{\mathrm{2}{n}−\mathrm{4}} \right)+\left(−\mathrm{1}\right)^{{n}} \left({U}_{\mathrm{2}{n}} +{U}_{\mathrm{2}{n}−\mathrm{2}} \right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:\Rightarrow\left(−\mathrm{1}\right)^{{n}} \:{U}_{\mathrm{2}{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}} =\left(−\mathrm{1}\right)^{{n}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}} \\ $$

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

let find U_(2n+1) we have U_(2n+1) =(1/(2n)) −U_(2n−1) ⇒ U_(2n+1) +U_(2n−1) =(1/(2n)) ⇒ Σ_(k=1) ^n (−1)^k (U_(2k+1) +U_(2k−1) )=Σ_(k=1) ^n (((−1)^k )/(2k)) ⇒ −(U_3 +U_1 )+U_5 +U_3 +....+(−1)^n (U_(2n+1) +U_(2n−1) ) =Σ_(k=1) ^n (((−1)^k )/(2k)) ⇒ U_(2n+1) =(((−1)^n )/2)Σ_(k=1) ^n (((−1)^k )/k) .

$${let}\:{find}\:{U}_{\mathrm{2}{n}+\mathrm{1}} \:\:\:{we}\:{have}\:{U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:−{U}_{\mathrm{2}{n}−\mathrm{1}} \:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}+\mathrm{1}} \:+{U}_{\mathrm{2}{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left({U}_{\mathrm{2}{k}+\mathrm{1}} +{U}_{\mathrm{2}{k}−\mathrm{1}} \right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}}\:\Rightarrow \\ $$$$−\left({U}_{\mathrm{3}} +{U}_{\mathrm{1}} \right)+{U}_{\mathrm{5}} +{U}_{\mathrm{3}} \:+....+\left(−\mathrm{1}\right)^{{n}} \left({U}_{\mathrm{2}{n}+\mathrm{1}} \:+{U}_{\mathrm{2}{n}−\mathrm{1}} \right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}}\:\Rightarrow \\ $$$${U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:. \\ $$

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

we see that (U_n ) is divergent but we have (−1)^n U_(2n) =Σ_(k=1) ^n (((−1)^k )/(2k−1)) =_(k−1=p) Σ_(p=0) ^(n−1) (((−1)^(p+1) )/(2p+1)) =−Σ_(p=0) ^(n−1) (((−1)^p )/(2p+1)) →−(π/4)(n→+∞) also (−1)^n U_(2n+1) =(1/2) Σ_(k=1) ^n (((−1)^k )/k) →−((ln(2))/2)(n→+∞)

$${we}\:{see}\:{that}\:\left({U}_{{n}} \right)\:{is}\:{divergent}\:\:{but}\:{we}\:{have} \\ $$$$\left(−\mathrm{1}\right)^{{n}} \:{U}_{\mathrm{2}{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}−\mathrm{1}}\:=_{{k}−\mathrm{1}={p}} \:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}} \\ $$$$=−\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}}\:\rightarrow−\frac{\pi}{\mathrm{4}}\left({n}\rightarrow+\infty\right) \\ $$$${also}\:\left(−\mathrm{1}\right)^{{n}} \:{U}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\rightarrow−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\left({n}\rightarrow+\infty\right) \\ $$

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

I_n =∫tan^(n−2) t×tan^2 tdt =∫tan^(n−2) t(sec^2 t−1)dt =∫tan^(n−2) tsec^2 tdt−∫tan^(n−2) tdt =∫tan^(n−2) t×d(tant)−I_(n−2) =(((tant)^(n−2+1) )/((n−2+1)))−I_(n−2) so U_n =∣(((tant)^(n−1) )/(n−1))∣_0 ^(π/4) −U_(n−2) U_n =(1/(n−1))−U_(n−2)

$${I}_{{n}} =\int{tan}^{{n}−\mathrm{2}} {t}×{tan}^{\mathrm{2}} {tdt} \\ $$$$=\int{tan}^{{n}−\mathrm{2}} {t}\left({sec}^{\mathrm{2}} {t}−\mathrm{1}\right){dt} \\ $$$$=\int{tan}^{{n}−\mathrm{2}} {tsec}^{\mathrm{2}} {tdt}−\int{tan}^{{n}−\mathrm{2}} {tdt} \\ $$$$=\int{tan}^{{n}−\mathrm{2}} {t}×{d}\left({tant}\right)−{I}_{{n}−\mathrm{2}} \\ $$$$=\frac{\left({tant}\right)^{{n}−\mathrm{2}+\mathrm{1}} }{\left({n}−\mathrm{2}+\mathrm{1}\right)}−{I}_{{n}−\mathrm{2}} \\ $$$${so}\:{U}_{{n}} =\mid\frac{\left({tant}\right)^{{n}−\mathrm{1}} }{{n}−\mathrm{1}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −{U}_{{n}−\mathrm{2}} \\ $$$$\boldsymbol{{U}}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−{U}_{{n}−\mathrm{2}} \\ $$$$ \\ $$

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