What is the value of $\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{128}+...}}}}$?

Question

An interesting radical I came up with and I'd like to know your approach.

Each term is of the form: $\frac{2^{2^r}}{2^{2^{r}+2^{r-1}+...+2+1}} = \frac{1}{2^{2^{r-1}+...+2+1}}$

Answer 1

If you multiply the expression by $1/\sqrt{2}$ and push it inside each successive radical, you get the second radical. So you have $x = \sqrt{1+x/\sqrt{2}}.$ When you solve for $x$, you get $x=\sqrt{2}.$

Answer 2

Let the number be $x$. Then

\begin{align} \sqrt{2}(x^2-1)&=x\\ 16x^2-8\sqrt{2}x-16&=0\\ (4x-\sqrt{2})^2&=18\\ 4x-\sqrt{2}&=\pm3\sqrt{2}\\ 4x&=4\sqrt{2}\qquad (\text{as }x\text{ is positive})\\ x&=\sqrt{2} \end{align}