The value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$?

First of all let's assume the series is convergent. Looking for fixed points we have:

$$x=\sqrt{1-\sqrt{1+x}}$$

Now we will try to solve this equation. First squaring both sides:

$$1-x^2=\sqrt{1+x} \\ \left(\left( 1-x\right)\left( 1+x\right) \right)^2=1+x$$

Note that $x$ must be nonnegative, therefore:

$$(1-x)^2(1+x)-1=0 \\ \Rightarrow x^3-x^2-x=0$$

So $x=0$ is a solution. The other solutions are:

$$x^2-x-1=0 \Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$$

where only $x=\frac{1 + \sqrt{5}}{2}$ is greater than or equal to zero and may look valid. But as people pointed out, one has to check if the answers actually fit into the initial equation. In this case $\frac{1 + \sqrt{5}}{2}$ doesn't, therefore the only fixed point we have found is $x=0$.

But $x=0$ cannot be the convergence limit(it doesn't converge smoothly). Assume we deflect $x=0$ with the tiny amount of $\epsilon$(or rather starting with a tiny $x_1=\epsilon$). Putting it back into our initial equations and getting the next $x$:

$$x_2=\sqrt{1-\sqrt{1+\epsilon}}\approx \sqrt{1-\left( 1+\frac{\epsilon}{2}\right)} \approx \frac{i\sqrt{\epsilon}}{\sqrt{2}}$$

Now for $\epsilon < \frac{1}{2}$, $|x_2|>|x_1|$; ergo $x=0$ cannot be the convergence limit. We have proved that this infinite radicals doesn't have a single limit.

The iteration of function $f(x) = \sqrt{1-\sqrt{1+x}}$ starting at $x=1$ approaches a 2-cycle of $.2229859448+.4133637969 i$ and $.2229859448-.4133637969 i$.