Convergence of alternating nested radicals Last evening, after reading a couple of questions about nested radicals, I started to wonder about problems involving what I will term "alternating nested radicals;" below is an example, which I found here. 

Prove the convergence of, and evaluate, $\sqrt{7 -\sqrt{7 + \sqrt{7...}}}$

It turns out that this nested radical converges to $2$, and this is not especially hard to argue on an ad-hoc basis. However, I became interested in the general solution of the problem of evaluating $\sqrt{q -\sqrt{q + \sqrt{q...}}} \text{ }\text{ }$ for an arbitrary positive real $q$. Despite some searching, I was not able to find a paper which stated a theorem on this, so I resorted to working it out for myself.
I have developed an argument which I believe gives the correct result for all $q > 1.$ I have checked its predictions against several alternating nested radicals, and they agree with computation. 
Theorem: $\sqrt{q -\sqrt{q + \sqrt{q...}}} = \frac{\sqrt{1 + 4(q-1)}-1}{2}$
Argument: We make two significant assumptions:


*

*that $\sqrt{q -\sqrt{q + \sqrt{q...}}} \text{ }\text{ }$ converges; and,

*inspired by the fact that $\sqrt{q +\sqrt{q + \sqrt{q...}}} \text{ }\text{ } = x$ satisfies $x^2 - x - q = 0$, we assume that $\sqrt{q -\sqrt{q + \sqrt{q...}}} \text{ }\text{ } = x$ satisfies $x^2 + x - a$ for some $a$.   


By the self-similarity of the alternating nested radical, we find 
\begin{align} 
\sqrt{q - \sqrt{q + x}} &= x \\
 q - \sqrt{q + x} &= x^2 \\
&= a-x \\
\therefore q + x - \sqrt{q + x} &= a
\end{align}
Solving this as a quadratic in $\sqrt{q+x}$ yields $$\sqrt{q+x} = \frac{\sqrt{1 + 4a} + 1}{2}.$$ Again using our second assumption, we have $$x = \frac{\sqrt{1 + 4a} - 1}{2}.$$ But now 
$$ \sqrt{q+x} - x = \frac{\sqrt{1 + 4a} + 1}{2} - \frac{\sqrt{1 + 4a} - 1}{2} = 1$$
and so we get
$$ q+x = (1+x)^2$$
and thus 
$$ x^2 + x - (q-1) = 0.$$
Solving this quadratic for $x$ then gives the theorem.

I am pretty sure that my second assumption above would be impossible to defend; so, basically, I have the following questions:


*

*How can we rigorously prove that the alternating nested radical converges?

*How can we show, for a general $q$, that it converges to the value given in the theorem?
And one other thing: any general references to ubiquitous convergence theorems or techniques would be greatly appreciated!
 A: Consider the function $f(x) = \sqrt{q - \sqrt{q+x}}$.
If we don't want to allow square roots of negative numbers, we need $x \ge -q$ and $q \ge \sqrt{q+x}$, so $x \le q^2 - q$.  On the interval $[-q, q^2 - q]$, $f$ is easily seen to be decreasing, with $$f'(x) = -\frac{1}{4 \sqrt{q - \sqrt{q+x}} \sqrt{q+x}}$$
Note also that
 $f(-q) = \sqrt{q}$ and $f(q^2-q) = 0$.    $f$ has a unique fixed point on the interval as long as $q^2 - q \ge 0$, i.e. $q \ge 1$.  This fixed point turns out to be 
$$p = \dfrac{-1+\sqrt{4q-3}}{2}$$ 
Now $f'(p) = -1$ for $q = 5/4$.  When $q < 5/4$, $f'(p) < -1$ and the fixed point is unstable.  When $q > 5/4$, $f'(p) > -1$ and the fixed point is stable.
When that is true, there is some interval around $p$ such that the
recursion $x_{n+1} = f(x_n)$, started at any $x_0$ in the interval, tends to 
the limit $p$. That gives some justification to identifying 
$\sqrt{q - \sqrt{q + \sqrt{q - \sqrt{\ldots}}}}$ as $p$.   
A: Let $R_0 =S_0 =\sqrt{q}$ and define the recurrences
$$A_{n+1}=\sqrt{q + S_n}$$
and
$$S_{n+1}=\sqrt{q - A_n}.$$
Then your nested radical is simply $\lim_{n \to \infty}S_n$.
If $q>\sqrt{q + \sqrt{q}}$,
then you can show by induction that for all $n>0$,
$$0<S_n \leq \sqrt{q} \leq A_n \leq \sqrt{q + \sqrt{q}}.
$$
Now $\{A_n\}$ clearly converges (by calculus), and this can be used to show
 $\{S_n\}$ is Cauchy.
Assuming convergence, let $R=\lim_{n \to \infty} S_n$. To evaluate the limit, notice
$$
R = \sqrt{q - \sqrt{q+R}},
$$
the solutions of which are among the solutions of $(R^2-q)^2=q+R$. Solving polynomial for $R$, and eliminating the negative solutions, we have
$$
R= \frac{1 +  \sqrt{4q+1}}{2}
$$
or
$$
R= \frac{-1 +  \sqrt{4q-3}}{2}.
$$
So the limit must be one of these solutions, if it exists.
It sounds like you have some good reason for believing the correct value is 
$\frac{1}{2}(-1 +  \sqrt{4q-3})$, so I'll stop here.
