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Question Number 193951 by Erico last updated on 23/Jun/23
Prove that: lim_(n→+∞) (1/(n!))∫^( n) _( 0) t^n e^(−t) dt = (1/2)
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}!}\underset{\:\mathrm{0}} {\int}^{\:\mathrm{n}} {t}^{{n}} {e}^{−{t}} {dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by senestro last updated on 24/Jun/23
lim_(n→+∞) (1/(n!))∫_0 ^( n) t^n e^(−t) dt=lim_(n→+∞) −e^(−t) (1+(t/(1!))+(t^2 /(2!))+...+(t^n /(n!)))∣_0 ^n =lim_(n→+∞) −e^(−n) (1+(n/(1!))+(n^2 /(2!))+...+(n^n /(n!)))+lim_(n→+∞) e^(−0) (1+(0/(1!))+(0^2 /(2!))+...+(0^n /(n!))) =lim_(n→+∞) −e^(−n) (1+(n/(1!))+(n^2 /(2!))+...+(n^n /(n!)))+lim_(n→+∞) 1(1) =lim_(n→+∞) −e^(−n) e^n +lim_(n→+∞) 1 =lim_(n→+∞) −1+lim_(n→+∞) 1 =−1+1=0 hence the question is wrong lim_(n→+∞) (1/(n!))∫_0 ^( n) t^n e^(−t) dt=0 but lim_(n→+∞) (1/(n!))∫_0 ^( ∞) t^n e^(−t) dt=1 and lim_(n→+∞) (1/(2n!))∫_0 ^( ∞) t^n e^(−t) dt=(1/2)
$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\:{n}} {t}^{{n}} {e}^{−{t}} {dt}=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}−{e}^{−{t}} \left(\mathrm{1}+\frac{{t}}{\mathrm{1}!}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+…+\frac{{t}^{{n}} }{{n}!}\right)\mid_{\mathrm{0}} ^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}−{e}^{−{n}} \left(\mathrm{1}+\frac{{n}}{\mathrm{1}!}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}!}+…+\frac{{n}^{{n}} }{{n}!}\right)+\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{e}^{−\mathrm{0}} \left(\mathrm{1}+\frac{\mathrm{0}}{\mathrm{1}!}+\frac{\mathrm{0}^{\mathrm{2}} }{\mathrm{2}!}+…+\frac{\mathrm{0}^{{n}} }{{n}!}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}−{e}^{−{n}} \left(\mathrm{1}+\frac{{n}}{\mathrm{1}!}+\frac{{n}^{\mathrm{2}} }{\mathrm{2}!}+…+\frac{{n}^{{n}} }{{n}!}\right)+\underset{{n}\rightarrow+\infty} {\mathrm{lim}1}\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}−{e}^{−{n}} {e}^{{n}} +\underset{{n}\rightarrow+\infty} {\mathrm{lim}1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}−\mathrm{1}+\underset{{n}\rightarrow+\infty} {\mathrm{lim}1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{1}+\mathrm{1}=\mathrm{0} \\ $$$${hence}\:{the}\:{question}\:{is}\:{wrong} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\:{n}} {t}^{{n}} {e}^{−{t}} {dt}=\mathrm{0} \\ $$$${but} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}!}\int_{\mathrm{0}} ^{\:\infty} {t}^{{n}} {e}^{−{t}} {dt}=\mathrm{1} \\ $$$${and} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}{n}!}\int_{\mathrm{0}} ^{\:\infty} {t}^{{n}} {e}^{−{t}} {dt}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by aba last updated on 24/Jun/23
lim_(n→∞) (1/(2n!))∫_0 ^∞ t^n e^(−t) dt=lim_(n→∞) ((Γ(n+1))/(2n!))=lim_(n→∞) ((n!)/(2n!))=(1/2) ✓
$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2n}!}\int_{\mathrm{0}} ^{\infty} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2n}!}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{n}!}{\mathrm{2n}!}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
Commented by Erico last updated on 26/Jun/23
Regarde : Posons f(t)= (t^n /(n!))e^(−t) ∀n∈N ∫^( 2n) _( n) f(t)dt=∫^( n) _( 0) f(2n−t)dt or ∫_(2n) ^( x) f(t)dt = ∫^( x−n) _( n) (((u+n))/(n!))e^(−u−n) du or u≥n ⇒ (((u+n)^n )/(n!))e^(−u−n) ≤ (((2u)^n )/(n!))e^(−u−n) ⇒Si x≥ 2n, ∫^( x−n) _( n) (((u+n)^n )/(n!))e^(−u−n) du ≤ ((2/e))^n ∫^( x−n) _( n) (u^n /(n!))e^(−u) du≤((2/e))^n ∫^( +∞) _( 0) (u^n /(n!))e^(−u) du ≤ ((2/e))^n donc lim_(n→+∞) (1/(n!))∫^( +∞) _( 2n) u^n e^(−u) du=0 ⇒ lim_(n→+∞) (1/(n!))∫^( 2n) _0 t^n e^(−t) dt=1 ⇒ lim_(n→+∞) [(1/(n!))∫^( n) _( 0) t^n e^(−t) dt+(1/(n!))∫^( 2n) _( n) t^n e^(−t) dt]=1 donc lim_(n→+∞) (1/(n!))∫^( n) _( 0) t^n e^(−t) dt ≠0
$$\mathrm{Regarde}\::\: \\ $$$$\mathrm{Posons}\:\mathrm{f}\left(\mathrm{t}\right)=\:\frac{\mathrm{t}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{t}} \: \\ $$$$\forall\mathrm{n}\in\mathbb{N}\:\:\:\:\:\underset{\:\mathrm{n}} {\int}^{\:\mathrm{2n}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{n}} \mathrm{f}\left(\mathrm{2n}−\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{or}\:\int_{\mathrm{2n}} ^{\:\mathrm{x}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\:=\:\underset{\:\mathrm{n}} {\int}^{\:\mathrm{x}−\mathrm{n}} \frac{\left(\mathrm{u}+\mathrm{n}\right)}{\mathrm{n}!}\mathrm{e}^{−\mathrm{u}−\mathrm{n}} \mathrm{du}\: \\ $$$$\mathrm{or}\:\mathrm{u}\geqslant\mathrm{n}\:\Rightarrow\:\frac{\left(\mathrm{u}+\mathrm{n}\right)^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{u}−\mathrm{n}} \:\leqslant\:\frac{\left(\mathrm{2u}\right)^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{u}−\mathrm{n}} \\ $$$$\Rightarrow\mathrm{Si}\:\mathrm{x}\geqslant\:\mathrm{2n},\:\:\:\:\:\:\underset{\:\mathrm{n}} {\int}^{\:\mathrm{x}−\mathrm{n}} \frac{\left(\mathrm{u}+\mathrm{n}\right)^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{u}−\mathrm{n}} \mathrm{du}\:\leqslant\:\left(\frac{\mathrm{2}}{\mathrm{e}}\right)^{\mathrm{n}} \underset{\:\mathrm{n}} {\int}^{\:\mathrm{x}−\mathrm{n}} \:\:\frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{u}} \mathrm{du}\leqslant\left(\frac{\mathrm{2}}{\mathrm{e}}\right)^{\mathrm{n}} \underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{u}^{\mathrm{n}} }{\mathrm{n}!}\mathrm{e}^{−\mathrm{u}} \mathrm{du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\:\left(\frac{\mathrm{2}}{\mathrm{e}}\right)^{\mathrm{n}} \\ $$$$\mathrm{donc}\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}!}\underset{\:\:\:\:\:\:\mathrm{2n}} {\int}^{\:\:\:\:+\infty} \mathrm{u}^{\mathrm{n}} \mathrm{e}^{−\mathrm{u}} \mathrm{du}=\mathrm{0} \\ $$$$\:\Rightarrow\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}!}\underset{\mathrm{0}} {\int}^{\:\mathrm{2n}} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}=\mathrm{1} \\ $$$$\Rightarrow\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{n}!}\underset{\:\mathrm{0}} {\int}^{\:\mathrm{n}} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}+\frac{\mathrm{1}}{\mathrm{n}!}\underset{\:\mathrm{n}} {\int}^{\:\mathrm{2n}} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}\right]=\mathrm{1} \\ $$$$\mathrm{donc}\:\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}!}\underset{\:\mathrm{0}} {\int}^{\:\mathrm{n}} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}\:\neq\mathrm{0} \\ $$$$ \\ $$

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