Question Number 40150 by maxmathsup by imad last updated on 16/Jul/18
$${let}\:{f}_{{n}} \left({x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{{n}} \right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} }\:\:\:{ddfined}\:{on}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}_{{n}} \rightarrow^{{cs}} \:{f}\:\left({n}\rightarrow+\infty\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}_{{n}} \left({x}\right){dx} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 21/Jul/18
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}_{{n}} \left({x}\right)\:=\left(\mathrm{1}+{x}^{{n}} \right)^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)} \:\:\:={e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+{x}^{{n}} \right)} \:\:\:\:{but} \\ $$$${ln}\left(\mathrm{1}+{x}^{{n}} \right)\sim\:{x}^{{n}} \:\:\:\left({n}\rightarrow+\infty\right)\:\Rightarrow−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+{x}^{{n}} \right)\:\sim−{x}^{{n}} \:−\frac{{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$${f}_{{n}} \left({x}\right)\:\sim\:\:{e}^{−{x}^{{n}} \:−\frac{{x}^{{n}} }{{n}}} \:\:\rightarrow^{{cs}} \:\:\:\:{f}\left({x}\right)=\mathrm{1}\:\:\left({n}\rightarrow+\infty\right) \\ $$