TITLE: Convergence of nested radicals
QUESTION [6 upvotes]: In general if we have $$\sqrt{a_0+\sqrt{a_1+\sqrt{a_2....}}}$$
 Are there easy ways of finding out if it converges, like from infinite products and series? Are there ways for converting the question of convergence of the radical to a one about series, like the logarithm does for products?

REPLY [3 votes]: According to the Herschfelds Convergence theorem, the sequence converges iff $a_n^{2^{-n}}$ converges, if $a_i>0  $. We can prove this by noting that, $$\sqrt{a_0+\sqrt{a_1+ \cdots + \sqrt{a_n}}}>{a_n^{2^{-n}}}$$
and also if $a_n<M^{2^{n}}$ then,
$$\sqrt{a_0+\sqrt{a_1+ \cdots + \sqrt{a_n}}}<\sqrt{M}\sqrt{1+\sqrt{1 + \cdots}}$$ which is known to converge, and the sequence is non decreasing, showing that it converges. Much thanks to muzzlator for pointing out the theorem.