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Let $S$ be the set of functions $f:\mathbb{R}\to \mathbb{R}$ such that $\sqrt{f(1)+\sqrt{f(2)+\sqrt{f(3)+\dots}}}$ converges.

A function $q(x)$ dominates $p(x)$ if there exist an m such that $q(x)\gt p(x)$ for all $x\gt m$.

Take all functions $f(x)$ from $S$ and put $O(f(x))$ in $S2$.

Which function $g(x)$ in $S2$ dominates all others?

Are there asymptotic lower and upper bounds on $g(x)$?

TROLLHUNTER
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  • Mark-up tip: use \to instead of -> – Arturo Magidin Mar 09 '11 at 06:49
  • Given a $g(x) \in S2$ doesn't $h(x)$=$g(x)$ if $x\ne 2$ or $g(2)+1$ if $x=2$ dominate it? – Ross Millikan Mar 09 '11 at 06:58
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    The question is about sequences $(f(n))_n$ rather than functions, and probably about nonnegative sequences. For every nonnegative sequence $(f(n))_n$ in $S$ and every positive $a$, $g(1)=1$ and $g(n)=af(n)$ for every $n\ge2$ defines a sequence $g$ in $S$. Hence $S$ can have no maximal element. – Did Mar 09 '11 at 09:00

1 Answers1

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Herschfeld proved that the nested radical converges iff $f_n^{2^{-n}}$ is bounded. In view of this, you consider bounded sequences under the same domination relation. It is clear that the increasing sequence of all integral constant sequences is cofinal (bounds every sequence). Coming back to the original problem, the following sequence is increasing and cofinal: $$ \begin{align*} &1,1,1,1,\ldots \\ &2,4,16,256,\ldots \\ &3,9,81,6561,\ldots \\ &4,16,256,65536,\ldots \\ &5,25,625,390625,\ldots \\ &\ldots \end{align*} $$ The $k$th sequence starts with $k$ and continues by repeated squaring.

Yuval Filmus
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