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!


2 Answers 2


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 $p$.


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.


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