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Question Number 66010 by jimful last updated on 07/Aug/19

$$Using\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}=e,provethat$$$$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{1}{n})}^n=e$$

Commented by Prithwish sen last updated on 07/Aug/19

$$=1+\mathrm{n}.\frac{1}{\mathrm{n}}+\mathrm{n}(\mathrm{n}-1)\frac{1}{2.{\mathrm{n}}^2}+\mathrm{n}(\mathrm{m}-1)(\mathrm{n}-2)\frac{1}{3!{\mathrm{n}}^3}+....$$$$=1+1+(1-\frac{1}{\mathrm{n}})\frac{1}{2!}+(1-\frac{1}{\mathrm{n}})(1-\frac{2}{\mathrm{n}})\frac{1}{3!}+...$$$$=1+1+\frac{1}{2!}+\frac{1}{3!}+....\mathrm{as}\mathrm{n}\to \mathrm{\infty }\Rightarrow \frac{1}{\mathrm{n}}\to 0$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{n}!}=\mathrm{e}\mathrm{please}\mathrm{check}.$$

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