Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k>0,k=1,2,\cdots)$ converge?

Question

I need to prove that if $$ \limsup\limits_{n\to\infty} \dfrac{\ln\ln a_n}{n}<\ln 2, $$ the sequence $\{t_n\}$ converge. (When $a_n\le 1$, $\ln\ln a_n$ is defined to be $-\infty$.)

I have no idea about this question. But if I consider the sequence $$ p_n^{(r)} = \sqrt[r]{a_1+\sqrt[r]{a_2+\cdots+\sqrt[r]{a_n}}}, r\in\mathbb{Z}_+$$ and in particular, when $r=1$, I have that $$ p_n^{(1)} = \sum_{k=1}^n a_k, $$ thus, if $$ \limsup\limits_{n\to\infty}\dfrac{\ln a_n}{n}<0(=\ln 1), $$ it's easy to obtain that $\{p_n^{(1)}\}$ converges. I guess there is a connection but I don't know how to continue.

Any hint?

I don't think there is a connection. (For one thing, nested cubic roots are obviously controlled by the same expression as square roots, just with a different constant.) Informally speaking, $a_n$ sits under $n$ roots, so if you want it to matter, it must grow at least as fast as $\exp(2^n)$. — Ivan Neretin, Oct 08 '20 at 17:26
Essentially a duplicate of Definition of convergence of a nested radical $\sqrt{a_1 + \sqrt{a_2 + \sqrt{a_3 + \sqrt{a_4+\cdots}}}}$?. See also Wikipedia article Nested Radical section on Herschfeld's Convergence Theorem. — Somos, Oct 08 '20 at 17:47

score 0 · Answer 1 · answered Oct 08 '20 at 17:50

This is equivalent to showing that there exists a constant $R$ such that $a_n\leq R^{2^n}$. To see this, first unwrap the $\sup$ condition you're given for large enough $n$ via:

$$\ln(\ln(a_n))/n\leq \ln(2)+\epsilon$$

for all $n\geq N$, giving $a_n\leq ce^{2^n}\leq R^{2^n}$.

Thus:

\begin{align*} \sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}}&\leq \sqrt{R^{2^1}+\sqrt{R^{2^2}+\cdots+\sqrt{R^{2^n}}}}\\ &= R\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}\\ &\leq 2R, \end{align*} and since the sequence is monotone increasing and bounded, we're done.

Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k>0,k=1,2,\cdots)$ converge?

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