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Question Number 211716 by liuxinnan last updated on 18/Sep/24

$$\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{1\times 2}+\frac{1}{1\times 2\times 3}+\cdot \cdot \cdot +\frac{1}{1\times 2\times 3\times \cdot \cdot \cdot \times x})=?$$

Answered by BHOOPENDRA last updated on 18/Sep/24

$$\underset{x\to \mathrm{\infty }}{\lim }(\underset{n=0}{\sum ^x}\frac{1}{n!}-1)$$$$x\to \mathrm{\infty }so$$$$S=(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}-1)$$$$\underset{x\to \mathrm{\infty }}{\lim }{S}_x=e-1(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e)$$

Commented by liuxinnan last updated on 19/Sep/24

$$why\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}=e$$

Commented by TonyCWX08 last updated on 19/Sep/24

$$UsingMaclaurinseries,$$$$f\left(x\right)=\frac{f\left(0\right)}{0!}{x}^0+\frac{f^{\prime }\left(0\right)}{1!}{x}^1+\frac{f^{″}\left(0\right)}{2!}{x}^2+\frac{f^{‴}\left(0\right)}{3!}{x}^3+...+\frac{f^n\left(0\right)}{n!}{x}^n$$$$$$$$letf\left(x\right)=e^x$$$$Nomatterhowmanytimesyoudifferentiateit,italwayswillbe{e}^x$$$$Hence,$$$$f^n\left(x\right)=e^x$$$$f^n\left(0\right)=1$$$$$$$$Substituteintotheseries,$$$$e^x=\frac{1}{0!}{x}^0+\frac{1}{1!}{x}^1+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+...$$$$letx=1$$$$e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}...$$$$$$$$Hence,e=\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n!}\right)$$

Commented by TonyCWX08 last updated on 19/Sep/24

$$Hopeyouunderstandnow,LiuXinNan$$

Commented by liuxinnan last updated on 19/Sep/24

$$Iget$$

Commented by BHOOPENDRA last updated on 19/Sep/24

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