The sequence of nested radicals $\sqrt {a+\sqrt{a+\sqrt{a+...}}}$ is defined by
$x_1=\sqrt a$, $x_{n+1}=\sqrt{a+x_n}$, where $a>0$.
Here are three questions:
(a) Show that $x=\displaystyle\lim_{n\to\infty} x_n=\frac{1+\sqrt{1+4a}}{2}$.
(b) Prove $\displaystyle x-x_n\sim \frac{C_a}{(2x)^n}$ for some $C_a>0.$
(c) For $a=2$, what is the value of $C_2$ in (b)?
My attempt:
(a) is easy. It is easy to prove $\{x_n\}$ is monotone and bounded, hence $x=\displaystyle\lim_{n\to\infty} x_n$ exists. Then we have the equation $x^2-x-a=0$.
For (b), I start by writing:
\begin{align} x-x_n &= \frac{x^2-x_n^2}{x+x_n} \\ &= \frac{(x+a)-(x_{n-1}+a)}{x+x_n} \\ &= \frac{x-x_{n-1}}{x+x_n} \end{align}
By induction,
$$x-x_n= \frac{x}{\prod_{k=1}^n (x+x_k)}$$
and I got stuck at this step.
For (c), once we prove $\displaystyle x_n=2\cos\left(\frac{\pi}{2^{n+1}}\right)$, it is not hard to see the constant $C_2$ is equal to $\displaystyle \frac{\pi^2}{4}$. This is a standard example of trigonometric substitution.
Any idea on how to tackle (b)?
Notation: $f(n)\sim g(n)$ if and only if $\displaystyle\lim_{n\to \infty} \frac{f(n)}{g(n)}=1$.
Let $x_n:=x-h_n$, so that $$(x-h_{n+1})^2=a+x-h_n$$ which simplifies to $$h_{n+1}^2-2xh_{n+1}+h_n=0$$ \begin{align*}h_{n+1}&=x-\sqrt{x^2-h_n}\\ &=\frac{h_n}{2x}(1+\frac{h_n}{4x^2}+\frac{h_n^2}{8x^4}+\cdots)\\ \end{align*}
So multiplying out gives $$h_n=\frac{h_1}{(2x)^n}\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^4}+\cdots)$$ so the required constant is $$C_a=h_1\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^4}+\cdots)$$
This constant need not have a closed formula but it exists by standard results on convergence of products.