Question Number 137482 by Algoritm last updated on 03/Apr/21
Answered by MJS_new last updated on 03/Apr/21
$$\mathrm{0} \\ $$
Commented by Algoritm last updated on 03/Apr/21
$$\mathrm{proof} \\ $$
Answered by TheSupreme last updated on 03/Apr/21
![∫_n ^(2n) (1/(2x^4 ))≤∫_n ^(2n) ((xdx)/(1+x^5 ))≤∫_n ^(2n) (1/x^4 )dx for n>1 we have x^5 >1>0 −(1/6)x^(−3) ∣_n ^(2n) ≤I≤−(1/3)x^(−3) ∣_n ^(2n) [−(1/(48n^3 ))+(1/(6n^3 ))]≤I≤[−(1/(24n^3 ))+(1/(3n^2 ))] (7/(48n^3 ))≤I_n ≤(7/(24n^3 )) for n→∞ 0≤I≤0→ I=0](Q137502.png)
$$\int_{{n}} ^{\mathrm{2}{n}} \frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} }\leqslant\int_{{n}} ^{\mathrm{2}{n}} \frac{{xdx}}{\mathrm{1}+{x}^{\mathrm{5}} }\leqslant\int_{{n}} ^{\mathrm{2}{n}} \frac{\mathrm{1}}{{x}^{\mathrm{4}} }{dx} \\ $$$${for}\:{n}>\mathrm{1}\:{we}\:{have}\:{x}^{\mathrm{5}} >\mathrm{1}>\mathrm{0} \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}{x}^{−\mathrm{3}} \mid_{{n}} ^{\mathrm{2}{n}} \leqslant{I}\leqslant−\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}} \mid_{{n}} ^{\mathrm{2}{n}} \\ $$$$\left[−\frac{\mathrm{1}}{\mathrm{48}{n}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{3}} }\right]\leqslant{I}\leqslant\left[−\frac{\mathrm{1}}{\mathrm{24}{n}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} }\right] \\ $$$$\frac{\mathrm{7}}{\mathrm{48}{n}^{\mathrm{3}} }\leqslant{I}_{{n}} \leqslant\frac{\mathrm{7}}{\mathrm{24}{n}^{\mathrm{3}} } \\ $$$${for}\:{n}\rightarrow\infty \\ $$$$\mathrm{0}\leqslant{I}\leqslant\mathrm{0}\rightarrow\:{I}=\mathrm{0} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 03/Apr/21
![let U_n =∫_n ^(2n) ((xdx)/(1+x^5 )) ⇒U_n =_(x=(1/t)) ∫_(1/n) ^(2/n) (1/(t(1+(1/t^5 ))))(−(dt/t^2 )) =−∫_(1/n) ^(2/n) (dt/(t^3 +(1/t^2 ))) =−∫_(1/n) ^(2/n) (t^2 /(1+t^5 ))dt we have 1+t^5 >1 ⇒ ∫_(1/n) ^(2/n) (t^2 /(1+t^5 ))dt <∫_(1/n) ^(2/n) t^2 dt =[(1/3)t^3 ]_(1/n) ^(2/n) =(1/3){(8/n^3 )−(1/n^3 )}→0(n→+∞) ⇒ lim_(n→+∞) U_n =0](Q137516.png)
$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{n}} ^{\mathrm{2n}} \:\:\frac{\mathrm{xdx}}{\mathrm{1}+\mathrm{x}^{\mathrm{5}} }\:\Rightarrow\mathrm{U}_{\mathrm{n}} =_{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}} \:\:\:\int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\frac{\mathrm{2}}{\mathrm{n}}} \:\frac{\mathrm{1}}{\mathrm{t}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{5}} }\right)}\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$=−\int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\frac{\mathrm{2}}{\mathrm{n}}} \:\:\:\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{3}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\:\:=−\int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\frac{\mathrm{2}}{\mathrm{n}}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{5}} }\mathrm{dt}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}+\mathrm{t}^{\mathrm{5}} \:>\mathrm{1}\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\frac{\mathrm{2}}{\mathrm{n}}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{5}} }\mathrm{dt}\:<\int_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\frac{\mathrm{2}}{\mathrm{n}}} \:\mathrm{t}^{\mathrm{2}} \:\mathrm{dt}\:=\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\mathrm{3}} \right]_{\frac{\mathrm{1}}{\mathrm{n}}} ^{\frac{\mathrm{2}}{\mathrm{n}}} =\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{8}}{\mathrm{n}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\right\}\rightarrow\mathrm{0}\left(\mathrm{n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} =\mathrm{0} \\ $$$$ \\ $$