Question Number 26359 by abdo imad last updated on 24/Dec/17
$${find}\:\:\:{lim}_{{n}−>\propto} \:\:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}−\mathrm{1}} {dt}\:\:\:. \\ $$
Commented by abdo imad last updated on 25/Dec/17
![let put I_n = ∫_0 ^n (1−(t/n))^(n−1) dt I_n = ∫_R f_n (t)dt / f_n (t) =(1−(t/n) )^(n−1) χ_([0,n]]) (t)dt due to (1−(t/n))^(n−1) = e^((n−1)ln(1−(t/n))) −−>_(n−>∝) e^(−t) f_n (t) c.s to f(t) =e^(−t ) plus that /f_(n(t)) /≤ e^(−t) ∀n by convergence dominee lim I_n = ∫_R _(n−>∝) lim f_n (x)dx = ∫_R e^(−t) χ_([[0,∝[) (t)dt = ∫_0 ^∞ e^(−t) dt = [ − e^(−t) ]_(t=0) ^∝ =1.](https://www.tinkutara.com/question/Q26474.png)
$${let}\:{put}\:{I}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}−\mathrm{1}} {dt} \\ $$$${I}_{{n}} \:\:=\:\:\:\:\int_{\mathbb{R}} \:\:{f}_{{n}} \left({t}\right){dt}\:\:\:\:\:\:/\:\:\:{f}_{{n}} \left({t}\right)\:\:=\left(\mathrm{1}−\frac{{t}}{{n}}\:\right)^{{n}−\mathrm{1}} \chi_{\left.\left[\mathrm{0},{n}\right]\right]} \:\left({t}\right){dt} \\ $$$${due}\:{to}\:\:\:\:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}−\mathrm{1}} =\:\:{e}^{\left({n}−\mathrm{1}\right){ln}\left(\mathrm{1}−\frac{{t}}{{n}}\right)} \:\:−−>_{{n}−>\propto} \:{e}^{−{t}} \\ $$$${f}_{{n}} \:\left({t}\right)\:{c}.{s}\:{to}\:{f}\left({t}\right)\:={e}^{−{t}\:\:} \:\:\:\:{plus}\:{that}\:\:\:/{f}_{{n}\left({t}\right)} \:\:/\leqslant\:{e}^{−{t}} \:\:\:\forall{n}\:\:{by}\:{convergence}\:{dominee} \\ $$$${lim}\:{I}_{{n}} \:\:=\:\:\int_{\mathbb{R}} \:_{{n}−>\propto} \:{lim}\:{f}_{{n}} \left({x}\right){dx}\:=\:\int_{\mathbb{R}} \:{e}^{−{t}} \:\chi_{\left[\left[\mathrm{0},\propto\left[\right.\right.\right.} \left({t}\right){dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {dt}\:\:=\:\left[\:−\:{e}^{−{t}} \right]_{{t}=\mathrm{0}} ^{\propto} =\mathrm{1}. \\ $$$$ \\ $$
Commented by abdo imad last updated on 25/Dec/17
![another method but easy let put (t/n)=x ⇒ I_n = ∫_0 ^1 (1−x)^(n−1 ) ndx = − [(1−x)^n ]_0 ^1 =1](https://www.tinkutara.com/question/Q26475.png)
$${another}\:{method}\:{but}\:{easy}\:\:\:{let}\:{put}\:\frac{{t}}{{n}}={x} \\ $$$$\Rightarrow\:\:\:{I}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}\:} {ndx}\:=\:−\:\left[\left(\mathrm{1}−{x}\right)^{{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:\:=\mathrm{1} \\ $$