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Question Number 30206 by .none. last updated on 18/Feb/18

$$\frac{1}{1-\frac{1}{1-\frac{1}{1-\frac{1}{1-x}}}}$$$$x=\frac{\sqrt{3}-1}{2}$$

Commented by abdo imad last updated on 18/Feb/18

$$letputa=1-\frac{1}{1-\frac{1}{1-x}}\Rightarrow N=\frac{1}{1-\frac{1}{a}}=\frac{1}{\frac{a-1}{a}}=\frac{a}{a-1}but$$$$a=1-\frac{1}{\frac{x}{x-1}}=1-\frac{x-1}{x}=\frac{1}{x}\Rightarrow N=\frac{\frac{1}{x}}{\frac{1}{x}-1}=\frac{1}{x(\frac{1}{x}-1)}=\frac{1}{1-x}$$$$withx=\frac{\sqrt{3}-1}{2}wegetN=\frac{1}{1-\frac{\sqrt{3}-1}{2}}=\frac{2}{2-\sqrt{3}+1}$$$$N=\frac{2}{3-\sqrt{3}}.$$

Answered by ajfour last updated on 18/Feb/18

$$1-\frac{1}{1-x}=\frac{x}{x-1}$$$$\Rightarrow 1-\frac{1}{1-\frac{1}{1-x}}=1-\frac{x-1}{x}=\frac{1}{x}$$$$\Rightarrow \frac{1}{1-\frac{1}{(1-\frac{1}{1-\frac{1}{x}})}}=\frac{1}{1-x}$$$$=\frac{1}{1-\left(\frac{\sqrt{3}-1}{2}\right)}=\frac{2}{3-\sqrt{3}}=\frac{2(3+\sqrt{3})}{6}$$$$=1+\frac{1}{\sqrt{3}}.$$

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