How can a divergent nested radical be regularized (analogously to Cesaro sum regularization of a divergent series)? The infinite series $1-1+1-1+\cdots$ diverges because the sequence of partial sums, $1,0,1,0,\ldots$ has no limit.  However, it is well know that we can get around this problem in a number of ways; the series is summable using alternate methods, such as the Cesaro sum
$$c_{n}=\frac{1}{n}\sum_{i=1}^{n}s_{n}=\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{i}a_{n}.$$
For the series mentioned above, the Cesaro summation gives $\frac{1}{2}$, which is also the analytic extension $\lim_{x\rightarrow-1}\frac{1}{1-x}$.  More elaborate summation methods can of course be used to regularize even trickier series.
However, I am having some trouble dealing with non-convergent sequences that do not arise as series sums, but which do still seem to have some "structure" that might be used to regularize them.  The real-valued sequence $1,0,1,0,\ldots$ arose as the partial sums in the example in the previous paragraph, but that same divergent sequence arises if you are trying to calculate a limit of
$$\sqrt{1-\sqrt{1-\sqrt{1-\sqrt{1-\cdots}}}}.$$
Becuase  $1,0,1,0,\ldots$ does not converge, this limit of nested radicals does not exist—but if it did exist, it seems that it ought to converge to $\phi^{-1}$, the inverse of the Golden Mean. $\phi$ itself is, of course, a similar limit
$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},$$
since this is clearly convergent and satisfies $\phi^{2}=1+\phi$.  The purported nested radical for $\phi^{-1}$ would obviously satisfy $\left(\phi^{-1}\right)^{2}=1-\phi^{-1}$, if only it converged.
So what I am asking is whether there is a technique for regularizing $\sqrt{1-\sqrt{1-\cdots}}$ to get $\phi^{-1}$ that can be expressed as some kind of (non-arithmetic) average over the sequence $1,0,1,0,\ldots$.  I think it is clear that one can concoct an ad hoc averaging procedure to produce an regularized version of that sequence that will converge to any value in $(0,1)$, but I am looking for one that would somehow encode the information that the sequence arose from a nested radical expression—so that it could, I hope, be extended to other similarly divergent nested radicals.
 A: One way to regularize the sequence $(x_n)_{n=1}^\infty=\sqrt{1-\sqrt{1-\sqrt{1-\sqrt{1-\cdots}}}}$ with $x_1=1$ is by varying the initial value $x_1$ infinitesimally, as follows.
We have $x_{n+1}=f(x_n)$ for natural $n$, where
$$f(x):=\sqrt{1-x}.$$
Let
$$c:=1/\phi=0.618\dots.$$
Note that $f(x)\in(c,1)$ for $x\in(0,c)$ and $f(x)\in(0,c)$ for $x\in(c,1)$.
Also, for
$$f_2(x):=f(f(x))=\sqrt{1-\sqrt{1-x}}$$
we have the following:
$$0<x<c\implies x<f_2(x)<c,$$
$$c<x<1\implies c<f_2(x)<x.$$
For each $s\in(c,1)$, consider now the sequence $(x_n(s))_{n=1}^\infty$ defined by the conditions
$$x_1(s)=s\quad\text{and}\quad x_{n+1}(s)=f(x_n(s))\quad\text{for natural $n$}.$$
Then it follows from above reasoning that
$$0<x_2(s)<x_4(s)<x_6(s)<\cdots<c<\cdots<x_5(s)<x_3(s)<x_1(s)<1.$$
So, the monotonic sequences $(x_{2k-1}(s))_{k=1}^\infty$ and $(x_{2k}(s))_{k=1}^\infty$ converge, and it is clear that the limit of both sequences is $c$, the only root in $(0,1)$ of the equations $x=f_2(x)$ and $x=f(x)$. So, $x_n(s)\to c$ for each $s\in(c,1)$. So, the generalized limit of the original sequence $(x_n)_{n=1}^\infty$ is
$$\lim_{s\uparrow1}\lim_{n\to\infty}x_n(s)=\lim_{s\uparrow1}c=c,$$
as desired.
