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.

• Instead of regularizing a series you can regularize a sequence (the sequence or partial sums of the series). Similarly, your divergent nested radical corresponds to a limit of a sequence of finite nested radicals. Thus, all the methods used on divergent series can be used here, also. Hardy's book, Divergent Series is the classic compendium of this material. Dec 22, 2020 at 15:01
• @GeraldEdgar Yes, but if I used the same averaging methods used for divergent sequences of partial sums, I would find that the regularized limit of $1,0,1,0,\ldots$ would be $\frac{1}{2}$, which is not the meaningful answer when that sequence arises out of the nested radicals.
– Buzz
Dec 22, 2020 at 15:03
• Do you have a response to the answer below? Aug 14, 2022 at 16:54

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: $$0 $$c
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 $$0 So, the monotonic sequences $$(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.