TITLE: Speed of convergence of continued radicals with constant term
QUESTION [5 upvotes]: This is an old prelim problem in Analysis.
Continued radicals of the form$$x_n=\sqrt{a+x_{n-1}},\qquad x_0=0$$have been considered in MSE before. It is easy to check that $x_n$ defines a bounded monotone increasing sequence and that $x_n\xrightarrow{n\to \infty}\ell _a,$ where $\ell _a$ is the unique positive solution to the quadratic equation $$x^2-x-a=0,$$ namely,$$\ell _a=\frac{1+\sqrt{1+4a}}{2}>1.$$ This is typically written as$$\ell _a=\sqrt{a+\sqrt{a+\ldots +\sqrt{a+\ldots }}}.$$ Since $$\ell _a-x_n=\frac{\ell _a^2-x_n^2}{\ell _a+x_n}=\frac{\ell _a-x_{n-1}}{\ell _a+x_n},$$ we have that\begin{align*}
\ell _a-x_n & =(\ell _a-x_1)\prod \limits _{k=1}^n\frac{1}{\ell _a+x_k} \\
& \leq (\ell _a-x_1)\frac{1}{\ell^n _a}\tag{1}\label{main}.
\end{align*}
The problem is to improve the rate of convergence $|\ell _a-x_n|=O( \ell _a^n)$. In particular, to show that\begin{align*}\ell _a-x_n & \sim \frac{C_a}{(2\ell_a)^n}\tag{2}\label{better}
\end{align*}for some constant $C_a>0$.
From \eqref{main} we have that$$b_n:=(2\ell _a)^n(\ell _a-x_n)=(\ell _a-x_1)\prod \limits _{k=1}^n\frac{2\ell _a}{\ell _a+x_k}\leq (\ell _a-x_1)\prod \limits _{k=1}^n\frac{\ell _a}{x_k}.$$Since $\dfrac{2\ell _a}{\ell _a+x_n}>1$, $b_n$ is monotone increasing. The result would follow if for example $p_n:=\prod \limits _{k=1}^n\dfrac{\ell _a}{x_k}$ were bounded above (or better yet, if $p_n$ converged).
Any hints will be appreciated.

REPLY [2 votes]: Hint:
Notice that from your inequality (1),
$\sum_n(\ell_a-x_n)$ converges since

$0<x_n<\ell_a$, $n\geq1$,
$\ell_a>1$, which means that $\sum_n\ell^{-n}_a<\infty$.

This in turn, implies that $\sum_n\frac{\ell_a-x_n}{\ell_a}<\infty$. Hence, the infinite product
$$q:=\prod_n\frac{x_n}{\ell_a}=\prod_n\Big(1-\frac{\ell_a-x_n}{\ell_a}\Big)$$ converges absolutely (which guarantees that $q\neq0$).