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Question Number 26359 by abdo imad last updated on 24/Dec/17

find   lim_(n−>∝)   ∫_0 ^n  (1−(t/n))^(n−1) dt   .

$${find}\:\:\:{lim}_{{n}−>\propto} \:\:\int_{\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}−\mathrm{1}} {dt}\:\:\:. \\ $$

Commented byabdo 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.

$${let}\:{put}\:{I}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}−\mathrm{1}} {dt} \\ $$ $${I}_{{n}} \:\:=\:\:\:\:\int_{\mathbb{R}} \:\:{f}_{{n}} \left({t}\right){dt}\:\:\:\:\:\:/\:\:\:{f}_{{n}} \left({t}\right)\:\:=\left(\mathrm{1}−\frac{{t}}{{n}}\:\right)^{{n}−\mathrm{1}} \chi_{\left.\left[\mathrm{0},{n}\right]\right]} \:\left({t}\right){dt} \\ $$ $${due}\:{to}\:\:\:\:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}−\mathrm{1}} =\:\:{e}^{\left({n}−\mathrm{1}\right){ln}\left(\mathrm{1}−\frac{{t}}{{n}}\right)} \:\:−−>_{{n}−>\propto} \:{e}^{−{t}} \\ $$ $${f}_{{n}} \:\left({t}\right)\:{c}.{s}\:{to}\:{f}\left({t}\right)\:={e}^{−{t}\:\:} \:\:\:\:{plus}\:{that}\:\:\:/{f}_{{n}\left({t}\right)} \:\:/\leqslant\:{e}^{−{t}} \:\:\:\forall{n}\:\:{by}\:{convergence}\:{dominee} \\ $$ $${lim}\:{I}_{{n}} \:\:=\:\:\int_{\mathbb{R}} \:_{{n}−>\propto} \:{lim}\:{f}_{{n}} \left({x}\right){dx}\:=\:\int_{\mathbb{R}} \:{e}^{−{t}} \:\chi_{\left[\left[\mathrm{0},\propto\left[\right.\right.\right.} \left({t}\right){dt} \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {dt}\:\:=\:\left[\:−\:{e}^{−{t}} \right]_{{t}=\mathrm{0}} ^{\propto} =\mathrm{1}. \\ $$ $$ \\ $$

Commented byabdo 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

$${another}\:{method}\:{but}\:{easy}\:\:\:{let}\:{put}\:\frac{{t}}{{n}}={x} \\ $$ $$\Rightarrow\:\:\:{I}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{x}\right)^{{n}−\mathrm{1}\:} {ndx}\:=\:−\:\left[\left(\mathrm{1}−{x}\right)^{{n}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:\:=\mathrm{1} \\ $$

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