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.

  • $\begingroup$ 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. $\endgroup$ Dec 22, 2020 at 15:01
  • $\begingroup$ @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. $\endgroup$
    – Buzz
    Dec 22, 2020 at 15:03
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Aug 14, 2022 at 16:54

1 Answer 1


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.


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