Question Number 207634 by SANOGO last updated on 21/May/24
$${calculer}\:{lim}\:\:{n}\rightarrow+{oo}\:{f}_{{n}} \left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)=\int_{\mathrm{0}^{} } ^{+{oo}} \frac{{ne}^{−{x}} }{\mathrm{1}+{nx}}{dx}\:\:\:/{x}\in\left[\mathrm{0}+{oo}\left[\right.\right. \\ $$
Answered by Berbere last updated on 21/May/24
![i Think is U_n =∫_0 ^∞ ((ne^(−x) )/(1+nx))dx=^(IBP) [ln(1+nx)e^(−x) ]_0 ^∞ +∫_0 ^∞ ln(1+nx)e^(−x) dx =∫_0 ^1 ln(1+nx)e^(−x) +∫_1 ^∞ ln(1+nx)e^(−x) dx ∫_0 ^1 ln(1+nx)e^(−x) dx>0⇒U_n >∫_1 ^∞ ln(1+nx)e^(−x) dx ∀x≥1;ln(1+nx)≥ln(1+n)⇒U_n ≥∫_1 ^∞ ln(1+n)e^(−x) dx =ln(1+n)[−e^(−x) ]=((ln(1+n))/e) U_n >((ln(1+n))/e)⇒U_n →+∞](Q207635.png)
$${i}\:{Think}\:{is}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \frac{{ne}^{−{x}} }{\mathrm{1}+{nx}}{dx}\overset{{IBP}} {=}\left[{ln}\left(\mathrm{1}+{nx}\right){e}^{−{x}} \right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} {ln}\left(\mathrm{1}+{nx}\right){e}^{−{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{nx}\right){e}^{−{x}} +\int_{\mathrm{1}} ^{\infty} {ln}\left(\mathrm{1}+{nx}\right){e}^{−{x}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{nx}\right){e}^{−{x}} {dx}>\mathrm{0}\Rightarrow{U}_{{n}} >\int_{\mathrm{1}} ^{\infty} {ln}\left(\mathrm{1}+{nx}\right){e}^{−{x}} {dx} \\ $$$$\forall{x}\geqslant\mathrm{1};{ln}\left(\mathrm{1}+{nx}\right)\geqslant{ln}\left(\mathrm{1}+{n}\right)\Rightarrow{U}_{{n}} \geqslant\int_{\mathrm{1}} ^{\infty} {ln}\left(\mathrm{1}+{n}\right){e}^{−{x}} {dx} \\ $$$$={ln}\left(\mathrm{1}+{n}\right)\left[−{e}^{−{x}} \right]=\frac{{ln}\left(\mathrm{1}+{n}\right)}{{e}} \\ $$$${U}_{{n}} >\frac{{ln}\left(\mathrm{1}+{n}\right)}{{e}}\Rightarrow{U}_{{n}} \rightarrow+\infty \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by SANOGO last updated on 21/May/24
$${merci}\:\:{beaucoup} \\ $$