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Question Number 204879 by universe last updated on 09/Aug/24
lim_(n→∞) n!(e−x_n ) = ? where x_(n ) = 1+(1/(1!))+(1/(2!))+...+(1/(n!))
$$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}!\left({e}−\mathrm{x}_{\mathrm{n}} \right)\:=\:? \\ $$$$\:\:\mathrm{where}\:\mathrm{x}_{\mathrm{n}\:} =\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+…+\frac{\mathrm{1}}{\mathrm{n}!} \\ $$
Commented by mr W last updated on 01/Mar/24
you mean n→∞ ?
$${you}\:{mean}\:{n}\rightarrow\infty\:? \\ $$
Commented by universe last updated on 01/Mar/24
yes sir
$${yes}\:{sir} \\ $$
Answered by Frix last updated on 01/Mar/24
??? x_n =Σ_(k=0) ^n (1/(k!)) ⇒ x_0 =1 0!(e−x_0 )=1×(e−1)=e−1
$$??? \\ $$$${x}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}!}\:\Rightarrow\:{x}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{0}!\left(\mathrm{e}−{x}_{\mathrm{0}} \right)=\mathrm{1}×\left(\mathrm{e}−\mathrm{1}\right)=\mathrm{e}−\mathrm{1} \\ $$
Answered by witcher3 last updated on 02/Mar/24
e−x_n =Σ_(k≥n+1) (1/(k!))=(1/((n+1)!))Σ_(k≥n+1) (((n+1)!)/(k!))=(1/((n+1)!))(Σ_(k≥n+1) (((n+1)!)/(k!))) Σ_(k≥n+1) (((n+1)!)/(k!))=(1+(1/((n+2)))+(1/((n+2)(n+3)))= Σ_(k≥n+1) (((n+1)!)/(k!))≤1+Σ_(k≥n+2) (1/((n+2)^(k−n) )) ≤1+Σ_(k≥2) (1/2^k )=(3/2) ;k+1≥2;n≥1 e−x_n =(1/((n+1)!))Σ(((n+1)!)/(k!))≤(3/(2(n+1)!)) ⇒n!∣e−x_n ∣≤(3/(2(n+1))) 0≤lim_(n→∞) n!∣e−x_n ∣≤lim_(n→∞) (3/(2(n+1)))=0
$$\mathrm{e}−\mathrm{x}_{\mathrm{n}} =\underset{\mathrm{k}\geqslant\mathrm{n}+\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{k}!}=\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\underset{\mathrm{k}\geqslant\mathrm{n}+\mathrm{1}} {\sum}\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{k}!}=\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\left(\underset{\mathrm{k}\geqslant\mathrm{n}+\mathrm{1}} {\sum}\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{k}!}\right) \\ $$$$\underset{\mathrm{k}\geqslant\mathrm{n}+\mathrm{1}} {\sum}\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{k}!}=\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{3}\right.}\right)= \\ $$$$\underset{\mathrm{k}\geqslant\mathrm{n}+\mathrm{1}} {\sum}\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{k}!}\leqslant\mathrm{1}+\underset{\mathrm{k}\geqslant\mathrm{n}+\mathrm{2}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{k}−\mathrm{n}} }\:\leqslant\mathrm{1}+\underset{\mathrm{k}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{3}}{\mathrm{2}}\:\:;\mathrm{k}+\mathrm{1}\geqslant\mathrm{2};\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{e}−\mathrm{x}_{\mathrm{n}} =\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\Sigma\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{k}!}\leqslant\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)!} \\ $$$$\Rightarrow\mathrm{n}!\mid\mathrm{e}−\mathrm{x}_{\mathrm{n}} \mid\leqslant\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\mathrm{0}\leqslant\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\mathrm{n}!\mid\mathrm{e}−\mathrm{x}_{\mathrm{n}} \mid\leqslant\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}=\mathrm{0} \\ $$

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