# Convergence of (infinite) potence tower?

Let $(a_n)_{n\in\mathbb{N}}$ be a series in $\mathbb{C}$ or $\mathbb{R}$. Which contraints must $(a_n)$ match to make $b_n := a_1^{a_2^{...^{a_n}}}$ converge for $n\rightarrow\infty$?

For constant series with $e^{-e}<a_n<e^{1/e}$, the series $b_n$ converges; So I guess series converging toward a value within $(e^{-e},e^{1/e})$ make $b_n$ converge aswell (the first few $a_i$ in $a_1^{a_2^{a_3^{...}}}$ do not matter at some point?!)

Is this the constraint that is to be applied? That there exists an $N\in\mathbb{N}$ such that $\forall n>N: a_n\in(e^{-e},e^{1/e})$? Is there any literature about this? Potence towers are pretty interesting ;)

EDIT: The question has still not been solved. However, we were able to prove that my Idea was wrong (which was pretty easy): $$5^{1^{5^{1^...}}}$$ converges. (For any arbitrary number, 5 is just an example.) Further more, it has to be assumed that a maximum of 1 $a_n$ is equal to $0$ (for obvious reasons aswell)

When using logarithms to unify the Potwnce tower, using $$a^{b}=e^{\log{a}*b}$$ to reduce $a_1^{a_2^{a_3^{...}}}$ to $$e^{\log{a_1}*a_2^{a_3^{...}}}$$ and so on, it is required that the remaining term (strongly) converges to $0$ in order to keep the resulting term $e^{e^{e^{...}}}$ finite.

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What is $b_1$ and does your recursion really express what you want to know, since it doesn't fit the title of the question? – WimC Dec 6 '12 at 17:49
Oh right, forgot. Fixed. – CBenni Dec 6 '12 at 17:53

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There is an article of D.F.Barrow (1936) "Infinite exponentials" in the "American Mathematical Monthly" Vol 43, No.3, Pg150-160, where he deals exactly with this question and uncovers some properties of such infinite nonconstant exponential-towers. Unfortunately it is not openly online, but via JStore one can access it. – Gottfried Helms Dec 7 '12 at 20:04

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– Gottfried Helms Dec 7 '12 at 20:12

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I fiddle with "tetration" for several years now. A lot of literature comes across that time... Even online-correspondents give sometimes hints :-) – Gottfried Helms Dec 7 '12 at 22:25
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