Tetration Asymptotic Series

[Catullus]

I've always thought we should write \(\eta_-\) for \(e^{-e}\), as it acts kinda like the polar opposite of \(\eta\).

How about an upside down eta?

So I'd bet it's something like:

\[
\eta_-\uparrow\uparrow n = e^{-1} + O(1/\sqrt{n})
\]

EDIT:

Found it in Milnor. If \(f(z) = e^{2\pi i/k}z + O(z^2)\), then in the attracting petals (there will be \(k\)), the function \(f^{\circ n}(z) = O(1/\sqrt[k]{n})\) So yes, an initial estimate would be \(O(\frac{1}{\sqrt{n}})\).

You did not rationalize your denominators.