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Question Number 211716 by liuxinnan last updated on 18/Sep/24
lim_(x→∞) (1+(1/(1×2))+(1/(1×2×3))+∙∙∙+(1/(1×2×3×∙∙∙×x)))=?
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\centerdot\centerdot\centerdot×{x}}\right)=? \\ $$
Answered by BHOOPENDRA last updated on 18/Sep/24
lim_(x→∞) (Σ_(n=0 ) ^x (1/(n!))−1) x→∞ so S=(Σ_(n=0) ^∞ (1/(n!))−1) lim_(x→∞) S_x =e−1 (Σ_(n=0) ^∞ (1/(n!))=e)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{n}=\mathrm{0}\:} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{n}!}−\mathrm{1}\right) \\ $$$${x}\rightarrow\infty\:{so}\: \\ $$$${S}=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}!}−\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{S}_{{x}} ={e}−\mathrm{1}\:\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}={e}\right) \\ $$
Commented by liuxinnan last updated on 19/Sep/24
why Σ_(n=0) ^∞ (1/(n!))=e
$${why}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}={e} \\ $$
Commented by TonyCWX08 last updated on 19/Sep/24
Using Maclaurin series, f(x)=((f(0))/(0!))x^0 +((f′(0))/(1!))x^1 +((f′′(0))/(2!))x^2 +((f′′′(0))/(3!))x^3 +...+((f^n (0))/(n!))x^n let f(x) = e^x No matter how many times you differentiate it, it always will be e^x Hence, f^n (x)=e^x f^n (0)=1 Substitute into the series, e^x =(1/(0!))x^0 +(1/(1!))x^1 +(1/(2!))x^2 +(1/(3!))x^3 +... let x =1 e=(1/(0!))+(1/(1!))+(1/(2!))+(1/(3!))... Hence, e=Σ_(n=0) ^∞ ((1/(n!)))
$${Using}\:{Maclaurin}\:{series}, \\ $$$${f}\left({x}\right)=\frac{{f}\left(\mathrm{0}\right)}{\mathrm{0}!}{x}^{\mathrm{0}} +\frac{{f}'\left(\mathrm{0}\right)}{\mathrm{1}!}{x}^{\mathrm{1}} +\frac{{f}''\left(\mathrm{0}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{f}'''\left(\mathrm{0}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…+\frac{{f}^{{n}} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} \\ $$$$ \\ $$$${let}\:{f}\left({x}\right)\:=\:{e}^{{x}} \\ $$$${No}\:{matter}\:{how}\:{many}\:{times}\:{you}\:{differentiate}\:{it},\:{it}\:{always}\:{will}\:{be}\:{e}^{{x}} \\ $$$${Hence},\: \\ $$$${f}^{{n}} \left({x}\right)={e}^{{x}} \\ $$$${f}^{{n}} \left(\mathrm{0}\right)=\mathrm{1} \\ $$$$ \\ $$$${Substitute}\:{into}\:{the}\:{series}, \\ $$$${e}^{{x}} =\frac{\mathrm{1}}{\mathrm{0}!}{x}^{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{1}!}{x}^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!}{x}^{\mathrm{3}} +… \\ $$$${let}\:{x}\:=\mathrm{1} \\ $$$${e}=\frac{\mathrm{1}}{\mathrm{0}!}+\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}… \\ $$$$ \\ $$$${Hence},\:{e}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}!}\right) \\ $$
Commented by TonyCWX08 last updated on 19/Sep/24
Hope you understand now, Liu Xin Nan
$${Hope}\:{you}\:{understand}\:{now},\:{Liu}\:{Xin}\:{Nan} \\ $$
Commented by liuxinnan last updated on 19/Sep/24
I get
$${I}\:{get} \\ $$
Commented by BHOOPENDRA last updated on 19/Sep/24
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