Question Number 66344 by mathmax by abdo last updated on 12/Aug/19
![let f_n (x)=(1/((1+x^n )^(1+(1/n)) )) defined on [0,1] 1)prove that f_n →^(cs) to a function f on[0,1] 2) calculate I_n =∫_0 ^1 f_n (x)dx](https://www.tinkutara.com/question/Q66344.png)
$${let}\:{f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{{n}} \right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} }\:\:\:{defined}\:{on}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:{f}_{{n}} \rightarrow^{{cs}} \:\:{to}\:{a}\:{function}\:{f}\:{on}\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
![1) ona f_n (0) =1 and f_n (1) =(1/2^(1+(1/n)) ) →(1/2) (n→+∞) and for 0<x<1 f_n (x) =(1+x^n )^(−(1+(1/n))) =e^(−(1+(1/n))ln(1+x^n )) we have x^n →0 ⇒ ln(1+x^n )∼x^n ⇒−(1+(1/n))ln(1+x^n )∼−(1+(1/n))x^n ⇒ f_n (x)∼e^(−(1+(1/n))x^(n ) ) ⇒ f_n ^( cs) → 1 on]0,1[](https://www.tinkutara.com/question/Q66811.png)
$$\left.\mathrm{1}\right)\:{ona}\:{f}_{{n}} \left(\mathrm{0}\right)\:=\mathrm{1}\:{and}\:{f}_{{n}} \left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} }\:\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:\left({n}\rightarrow+\infty\right)\:{and}\:{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${f}_{{n}} \left({x}\right)\:=\left(\mathrm{1}+{x}^{{n}} \right)^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)} \:={e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+{x}^{{n}} \right)} \:\:\:{we}\:{have}\:{x}^{{n}} \:\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}^{{n}} \right)\sim{x}^{{n}} \:\Rightarrow−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+{x}^{{n}} \right)\sim−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){x}^{{n}} \:\Rightarrow \\ $$$$\left.{f}_{{n}} \left({x}\right)\sim{e}^{−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){x}^{{n}\:} } \:\Rightarrow\:\:{f}_{{n}} ^{\:\:\:\:\:{cs}} \rightarrow\:\mathrm{1}\:\:\:{on}\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$