Convergence of $\sqrt{1-\sqrt{{1}-\sqrt{{1}-\sqrt{{1-\sqrt{{1}}}...}}}}$

Question

Recently, as is evident from many of my recent questions, I have been very interested in nested radicals. Recently I attempted to investigate the following infinite nested radical and arrived at a strange, counter-intuitive result. $$\sqrt{1-\sqrt{{1}-\sqrt{{1}-\sqrt{{1-\sqrt{{1...}}}}}}}$$ At first glance, it would seem that the value of the nested radical must be either $0$ or $1$. However, if we make the assumption that a single value can be assigned to it, ie it converges, we arrive at a different value for it: Let $$x=\sqrt{1-\sqrt{{1}-\sqrt{{1}-\sqrt{{1-\sqrt{{1...}}}}}}}\implies x=\sqrt{1-x}\implies x^2=1-x \implies x^2+x-1=0$$ Using the quadratic formula, we get $$x=\frac{-1+\sqrt5}{2}$$ as $x$ is necessarily positive. Can this result be true? Or is my assumption that allowed me to label the radical as $x$ and assume it converges incorrect?

Answer 1

Whenever we find $\ldots$ in a formula we must be clear what they are meaning. My interpreation of a formula like $$\sqrt{1-\sqrt{1-\sqrt{1-\sqrt{1-\ldots}}}}$$ is that this is the limit of the sequence $$x_1=\sqrt{1}\\ x_2=\sqrt{1-\sqrt{1}}\\ x_3=\sqrt{1-\sqrt{1-\sqrt{1}}}\\ \ldots\\ x_{n+1}=\sqrt{1-x_n}$$

If this is the meaning of this expression then we have

$$x_1=1\\ x_2=0\\ x_3=1\\ x_4=0\\ x_5=1\\ x_6=0\\ \ldots $$

and this sequence does not converge at all.

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Your calculation says:

If $x_n$ converges, then the limit will be $\frac{-1+\sqrt5}{2}$ or $\frac{-1-\sqrt5}{2}$

This is true. But $x_n$ does not converge, so this result is useless.