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

p r o v e t h a t

\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}} = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + ⋱}}}

Answered by Olaf_Thorendsen last updated on 17/Jul/21

φ^{2} = φ + 1 (g o l d e n r a t i o \frac{1 + \sqrt{5}}{2})

\Rightarrow φ = 1 + \frac{1}{φ}

φ = 1 + \frac{1}{1 + \frac{1}{φ}}

φ = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{φ}}}

\dots e t c \dots

φ = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\dots}}}

Let x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}}

x^{2} = 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}

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}

But \frac{1 - \sqrt{5}}{2} < 0 : impossible because x > 1

Finally x = \frac{1 + \sqrt{5}}{2} = φ