convergence of $u_n=\sqrt{a_1+\sqrt{a_2+\sqrt{\dots+\sqrt{a_n}}}}$

Question

Let $(a_n)_{n\in\mathbb{N}}$ be a real strictly positive sequence, and $\forall n\in\mathbb{N},u_n=\sqrt{a_1+\sqrt{a_2+\sqrt{\dots+\sqrt{a_n}}}}$

I have shown that for $a_n=1$, $u_n$ converges to $\phi=\frac{1+\sqrt{5}}2$ and that for $a_n=\lambda^{2^n}$,$u_n$ converges to $\lambda\cdot\phi$.

I now have to show that $u_n$ converges iff $\ln a_n=\mathcal{O}(2^n)$.

I have managed to show that $\ln a_n=\mathcal{O}(2^n)\Rightarrow$$u_n$ converges.

How do i show the other way of the equivalence ?

PS : if I have made a mistake in the statements I think I have proved above, please tell me !

Answer 1

Since the sequence $u_n$ is increasing, it converges if and only if it's bounded above.

You said you've already proved one direction, which follows from the calculation you did for $a_n = \lambda^{2^n}$. The remaining direction assumes $u_n$ converges and asks for an estimate of $a_n$. Therefore, say $u_n \leq M$ is an upper bound.

We have $(a_n)^{1/2^n} = \sqrt{\sqrt{\sqrt{\dots \sqrt{a_n}}}} \leq u_n \leq M$.

From this it follows that $\log a_n \leq 2^n \log M$.