Question Number 61660 by maxmathsup by imad last updated on 05/Jun/19
$${let}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}} }\:{dt}\:\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} } \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{serie}\:\Sigma{ln}\left(\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\right)\:\:{and}\:{prove}\:\:{that}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} =\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 07/Jun/19
![1) we have U_n =∫_0 ^∞ ((1+t^3 )/((1+t^3 )^(n+1) )) dt =∫_0 ^∞ (dt/((1+t^3 )^(n+1) )) +∫_0 ^∞ (t^3 /((1+t^3 )^(n+1) )) dt ∫_0 ^∞ (dt/((1+t^3 )^(n+1) )) =U_(n+1) ∫_0 ^∞ (t^3 /((1+t^3 )^(n+1) )) dt = (1/3)∫_0 ^∞ t (3t^2 )(1+t^3 )^(−n−1) dt by parts u =t and v^, =(3t^2 )(1+t^3 )^(−n−1) ⇒ ∫_0 ^∞ (t^3 /((1+t^3 )^(n+1) )) dt = (1/3){ [−(t/n)(1+t^3 )^(−n) ]_0 ^∞ +∫_0 ^∞ (1/n)(1+t^3 )^(−n) dt} =(1/(3n)) ∫_0 ^∞ (dt/((1+t^3 )^n )) =(1/(3n)) U_n ⇒ U_n = U_(n+1) +(1/(3n)) U_n ⇒(1−(1/(3n)))U_n =U_(n+1) ⇒ (((3n−1)/(3n)))U_n =U_(n+1) ⇒(U_(n+1) /U_n ) = 1−(1/(3n))](https://www.tinkutara.com/question/Q61723.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}+{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\:+\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\:{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\:={U}_{{n}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\:{dt}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:{t}\:\left(\mathrm{3}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{−{n}−\mathrm{1}} \:{dt}\:{by}\:{parts} \\ $$$${u}\:={t}\:\:\:\:{and}\:{v}^{,} \:=\left(\mathrm{3}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{−{n}−\mathrm{1}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{3}} }{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\:{dt}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{\:\:\left[−\frac{{t}}{{n}}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{−{n}} \right]_{\mathrm{0}} ^{\infty} \:+\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{−{n}} {dt}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}{n}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{{n}} }\:=\frac{\mathrm{1}}{\mathrm{3}{n}}\:{U}_{{n}} \:\Rightarrow \\ $$$${U}_{{n}} =\:{U}_{{n}+\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{3}{n}}\:{U}_{{n}} \:\Rightarrow\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}{n}}\right){U}_{{n}} ={U}_{{n}+\mathrm{1}} \:\:\Rightarrow \\ $$$$\left(\frac{\mathrm{3}{n}−\mathrm{1}}{\mathrm{3}{n}}\right){U}_{{n}} ={U}_{{n}+\mathrm{1}} \:\Rightarrow\frac{{U}_{{n}+\mathrm{1}} }{{U}_{{n}} }\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}{n}} \\ $$