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Question Number 204879 by universe last updated on 29/Feb/24

$$\underset{n\to 0}{\lim }\mathrm{n}!(e-{\mathrm{x}}_{\mathrm{n}})=?$$$$\mathrm{where}{\mathrm{x}}_{\mathrm{n}}=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{\mathrm{n}!}$$

Commented by mr W last updated on 01/Mar/24

$$youmeann\to \mathrm{\infty }?$$

Commented by universe last updated on 01/Mar/24

$$yessir$$

Answered by Frix last updated on 01/Mar/24

$$???$$$$x_n=\underset{k=0}{\sum ^n}\frac{1}{k!}\Rightarrow x_0=1$$$$0!(\mathrm{e}-x_0)=1\times (\mathrm{e}-1)=\mathrm{e}-1$$

Answered by witcher3 last updated on 02/Mar/24

$$\mathrm{e}-{\mathrm{x}}_{\mathrm{n}}=\sum _{\mathrm{k}⩾\mathrm{n}+1}\frac{1}{\mathrm{k}!}=\frac{1}{(\mathrm{n}+1)!}\sum _{\mathrm{k}⩾\mathrm{n}+1}\frac{(\mathrm{n}+1)!}{\mathrm{k}!}=\frac{1}{(\mathrm{n}+1)!}\left(\sum _{\mathrm{k}⩾\mathrm{n}+1}\frac{(\mathrm{n}+1)!}{\mathrm{k}!}\right)$$$$\sum _{\mathrm{k}⩾\mathrm{n}+1}\frac{(\mathrm{n}+1)!}{\mathrm{k}!}=(1+\frac{1}{(\mathrm{n}+2)}+\frac{1}{(\mathrm{n}+2)(\mathrm{n}+3})=$$$$\sum _{\mathrm{k}⩾\mathrm{n}+1}\frac{(\mathrm{n}+1)!}{\mathrm{k}!}⩽1+\sum _{\mathrm{k}⩾\mathrm{n}+2}\frac{1}{{(\mathrm{n}+2)}^{\mathrm{k}-\mathrm{n}}}⩽1+\sum _{\mathrm{k}⩾2}\frac{1}{2^{\mathrm{k}}}=\frac{3}{2};\mathrm{k}+1⩾2;\mathrm{n}⩾1$$$$\mathrm{e}-{\mathrm{x}}_{\mathrm{n}}=\frac{1}{(\mathrm{n}+1)!}\mathrm{\Sigma }\frac{(\mathrm{n}+1)!}{\mathrm{k}!}⩽\frac{3}{2(\mathrm{n}+1)!}$$$$\Rightarrow \mathrm{n}!\mid \mathrm{e}-{\mathrm{x}}_{\mathrm{n}}\mid ⩽\frac{3}{2(\mathrm{n}+1)}$$$$0⩽\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{n}!\mid \mathrm{e}-{\mathrm{x}}_{\mathrm{n}}\mid ⩽\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{3}{2(\mathrm{n}+1)}=0$$

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