does this nested radicle converge? $1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}$

Question

Let $a_0=1$ and $a_n=1+\sqrt{a_{n-1}}+\sqrt{1+\sqrt{a_{n-1}}}$

I want to know if the limit of $a_n$ as n goes to infinity. $$1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}}$$ When I tried to find a possible answer I tried to solve $1+\sqrt{x}+\sqrt{1+\sqrt{x}}=x$. I tried for an hour so I gave up and asked WolframAlpha and it got $x≈5.04891733952231$

I'm not sure how to find out if $a_n$ converges. I think it does

Answer 1

Notice the map $x\mapsto f(x) \stackrel{def}{=} 1 + \sqrt{x} + \sqrt{1 + \sqrt{x}}$ is defined and strictly increasing for $x \ge 0$.

Since $a_1 > a_0$, we have $$a_2 = f(a_1) > f(a_0) = a_1 \implies a_3 = f(a_2) > f(a_1) = a_2 \implies \cdots $$ So $a_n$ is an increasing sequence.

The map $f(x)$ has a fixed point at $b \sim 5.05$ (in fact, this is the first non-zero fixed point of $f(x)$). Since $a_0 < b$, we have $$a_1 = f(a_0) < f(b) = b \implies a_2 = f(a_1) < f(b) = b \implies \cdots$$ So $a_n$ is also bounded from above by $b$. As a result, $a_n$ converges to some value $a_\infty$ in $(a_1,b] = (1,b]$.

Now $f(x)$ is a continuous function, this implies

$$f(a_\infty) = f(\lim_{n\to\infty}a_n) = \lim_{n\to\infty} f(a_n) = \lim_{n\to\infty} a_{n+1} = a_\infty$$

So $a_\infty$ is also a fixed point of $f(x)$ too. Since $b$ is the first non-zero fixed point of $f(x)$, this forces $a_\infty = b$. ie. your sequence $a_n$ converges to the fixed point you have.

Answer 2

Starting from @Will Jagy first hint $$w^3 - 2 w^2 - w + 1 = 0\qquad \text{with} \qquad x=w^2$$ the cubic shows three real roots $(\Delta=49)$. Using the trigonometric method, they are $$w_k=\frac{2}{3} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \left(2 \pi k-\sec ^{-1}\left(2 \sqrt{7}\right)\right)\right)\right)\qquad \text{with} \qquad k=0,1,2$$ and the solution of concern is $w_0$ (not prooved).

This gives $$x=\frac{4}{9} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \sec ^{-1}\left(2 \sqrt{7}\right)\right)\right)^2$$ which is another expression of the result.