Question Number 66344 by mathmax by abdo last updated on 12/Aug/19
$${let}\:{f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{{n}} \right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} }\:\:\:{defined}\:{on}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:{f}_{{n}} \rightarrow^{{cs}} \:\:{to}\:{a}\:{function}\:{f}\:{on}\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
$$\left.\mathrm{1}\right)\:{ona}\:{f}_{{n}} \left(\mathrm{0}\right)\:=\mathrm{1}\:{and}\:{f}_{{n}} \left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} }\:\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:\left({n}\rightarrow+\infty\right)\:{and}\:{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${f}_{{n}} \left({x}\right)\:=\left(\mathrm{1}+{x}^{{n}} \right)^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)} \:={e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+{x}^{{n}} \right)} \:\:\:{we}\:{have}\:{x}^{{n}} \:\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}^{{n}} \right)\sim{x}^{{n}} \:\Rightarrow−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+{x}^{{n}} \right)\sim−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){x}^{{n}} \:\Rightarrow \\ $$$$\left.{f}_{{n}} \left({x}\right)\sim{e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){x}^{{n}\:} } \:\Rightarrow\:\:{f}_{{n}} ^{\:\:\:\:\:{cs}} \rightarrow\:\mathrm{1}\:\:\:{on}\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$