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Question Number 162026 by mnjuly1970 last updated on 25/Dec/21

$$$$ $$provethat....$$ $$$$ $${(1+\frac{1}{n})}^n<e<{(1+\frac{1}{n})}^{n+1}$$ $$$$ $$$$

Answered by mindispower last updated on 25/Dec/21

$$nottrue$$ $${(1+\frac{1}{n})}^{n+1}<{(1+\frac{1}{n})}^n,n>0$$

Commented byAr Brandon last updated on 25/Dec/21

$$n=1$$ $${\left(2\right)}^2=4$$ $${\left(2\right)}^1=2$$ $${(1+\frac{1}{n})}^{n+1}>{(1+\frac{1}{n})}^n\mathrm{for}n=1>0$$

Answered by Ar Brandon last updated on 25/Dec/21

$$\mathrm{From}{e}^x⩾x+1\mathrm{\forall }x>0$$ $$\Rightarrow \ln e^x⩾\ln (x+1)\Rightarrow x⩾\ln (x+1)$$ $$\Rightarrow \frac{1}{n}⩾\ln (1+\frac{1}{n})\Rightarrow 1⩾n\ln (1+\frac{1}{n})$$ $$\Rightarrow 1⩾\ln {(1+\frac{1}{n})}^n\Rightarrow e⩾{(1+\frac{1}{n})}^n...\left(1\right)$$ $${(1+\frac{1}{n})}^{n+1}⩾{(1+\frac{1}{n+1})}^{n+1}...\left(2\right)$$ $${(1+\frac{1}{n})}^{n+1}⩾{(1+\frac{1}{N})}^N=e,N=\frac{1}{n+1}$$ $$\left(1\right)\mathrm{and}\left(2\right)$$ $$\Rightarrow {(1+\frac{1}{n})}^{n+1}⩾e⩾{(1+\frac{1}{n})}^n\mathrm{\forall }n\in {\mathbb{N}}^{+}$$

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