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Question Number 147071 by alcohol last updated on 17/Jul/21

$$provethat$$$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots }}}$$

Answered by Olaf_Thorendsen last updated on 17/Jul/21

$${\phi }^2=\phi +1\left(goldenratio\frac{1+\sqrt{5}}{2}\right)$$$$\Rightarrow \phi =1+\frac{1}{\phi }$$$$\phi =1+\frac{1}{1+\frac{1}{\phi }}$$$$\phi =1+\frac{1}{1+\frac{1}{1+\frac{1}{\phi }}}$$$$...etc...$$$$\phi =1+\frac{1}{1+\frac{1}{1+\frac{1}{...}}}$$$$\mathrm{Let}x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$$$$x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}$$$$x^2=1+x$$$$x^2-x-1=0$$$$(x-\frac{1-\sqrt{5}}{2})(x-\frac{1+\sqrt{5}}{2})=0$$$$x=\frac{1\pm \sqrt{5}}{2}$$$$\mathrm{But}\frac{1-\sqrt{5}}{2}<0:\mathrm{impossible}\mathrm{because}x>1$$$$\mathrm{Finally}x=\frac{1+\sqrt{5}}{2}=\phi$$

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