Nature of $u_n(x)=\sqrt{a_1(x)+\sqrt{a_2(x)+\sqrt{\dots+\sqrt{a_n(x)}}}},a_n(x)=x^{2^n}$

Question

In a question posed in this forum convergence of $u_n=\sqrt{a_1+\sqrt{a_2+\sqrt{\dots+\sqrt{a_n}}}}$ , the author proved that if $a_n=\lambda^{2^n}$ then $a_n$ tends to $\phi\lambda,\phi=\frac{1+\sqrt{5}}{2}$. Now, if we take $a_n(x)=x^{2^n}$, does the sequence converges to $\phi x$ and how to show that? Thank you in advance.

Answer 1

I guess the OP wants to know why if $a_n=x^{2^n}$ the series $u_n$ conveges to $\phi x$, something that is not proven in the linked question but taken for granted.

To do this, consider the following steps. Take first $x=1$, so that $a_n=1$ for each $n$.

(1) When $x=1$, the recursive relation is true:

$u_0=1, u_{n+1}=\sqrt{1+u_n}$

Suppose a limit exists $l$. Than it will satisfy (why?) $l=\sqrt{1+l}$

(2) To prove the limit exists, consider visually the function $f(x)=\sqrt{1+x}$ for positive $x$ and its intersections with the line $y=x$. Can you plot the succession of points $u_n$ ? In order to formalize the intuition show that:

$0\le x\le\phi \rightarrow 0\le f(x)\le\phi $

This shows that the sequence is bounded. Since it is increasing, it has a limit.

(3) Now change to $x>1$. Let's call $u_n(x)$ the new $u_n$. Can you show that $u_n(x)=x u_n(1)$ at a fixed $n$ ? (check small $n$ to develop an intuition) Than just take the limit of this relation and exploit the result you already found for $x=1$.