Do infinitely nested radicals have any applications?

Question

There is a simple necessary and sufficient condition for a continued radical of the form $\sqrt{a_1 + \sqrt{a_2 + \dotsc}}$ to converge (where all terms $a_1, a_2$ etc. are nonnegative). Namely, that the sequence $n \mapsto a_n^{2^{-n}}$ should be bounded. This is known as Herschfeld's Convergence Theorem (though it was discovered independently of Herschfeld by Paul Wiernsberger thirty years before).

Lately, I've developed a constructive proof of this theorem, which you can find here.

It leads me to wonder: Is there any actual use for continued square roots, apart from studying them for their own sake? Googling doesn't show up anything.

Answer 1

The first application goes back to Archimedes. Let me explain how. $$ \underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n}\to 2. $$ The question is how fast. It turns out that the rate of convergence can be caught quite precisely from the formula: $$ 2^{n+2}\cdot\sqrt{2-{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}}\to 2\pi. $$ This formula has an easy geometric interpretation: the expression on the left hand side equals the circumference of the regular $2^{n+2}$-gon inscribed in the unit circle. In fact, Archimedes used this approach to find the approximate value of $\pi$. The method was mastered later by Ludolph van Ceulen who published in 1596 the first 20-decimals of $\pi$.

Answer 2

The following is not an answer.

In a comment to @PiotrHajlasz's answer, I say that a method from my paper shows that $2-\underbrace{\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}_{n} \leq 2^{2^{-n}}(2^{2^{-n}} - 1)$. You can find a description of this in the subsection Overview and Strategy of my proof. But in case it's not clear, I describe how to do this below:

The general idea is that given $U \geq L$ and a non-negative sequence $(a_i)_i$, we have that $$\begin{align} &\sqrt{a_1 + \sqrt{a_2 + \dotsb \sqrt{a_{n-1} + \sqrt{U}}}} - \sqrt{a_1 + \sqrt{a_2 + \dotsb \sqrt{a_{n-1} + \sqrt{L}}}} \leq \sqrt{0 + \sqrt{0 + \dotsc\sqrt{0 + \sqrt U}}} - \sqrt{0 + \sqrt{0 + \dotsc\sqrt{0 + \sqrt L}}} = U^{1/2^n} - L^{1/2^n}. \end{align}$$ In other words, we can upper bound the difference between an upper bound and lower bound by driving the terms $a_i$ down to $0$.

In the case of $\sqrt{2 + \sqrt{2 + \dotsb}}$, we have an upper bound in the form of $\sqrt{2 + \sqrt{2 + \dotsb \sqrt{2 + \sqrt{\color{red} 4}}}}=2$, and a lower bound in the form of $\sqrt{2 + \sqrt{2 + \dotsb \sqrt{2 + \sqrt{\color{red} 2}}}}$. The difference between upper and lower bound can thus be increased to $4^{1/2^n} - 2^{1/2^n} = 2^{1/2^n} (2^{1/2^n} - 1)$, which clearly goes to zero.