# How slow does the iteration $x_1=r$ , $x_{n+1}=r^{x_n}$ converge for $r=e^{-e}$?

The number $r:=e^{-e}$ is the smallest number for which the infinite power tower $r\uparrow r \uparrow r\uparrow \cdots$ converges. In other words, the iterarion $x_1=r$ , $x_{n+1}=r^{x_n}$ converges.

However, the convergence is rather slow.

Is there a tight asymptotic formula for the error we have if we take $n$ numbers $r$ ?

Experimentint with PARI/GP, it seems that the error is approximately $\frac{1}{\sqrt{n}}$

Yes, but depends upon the tightness you want? When you approach the fixed point $a=-1/\log r$ then you may compute the asymptotic behavior. Setting $f(x)=f_r(x)=r^x$ you have a pitchfork bifurcation for $g=f_r\circ f_r$ at $r=e^{-e}$ and $a=-1/\log r$. This leads to the asymptotic behavior: $$g(a+z) = a+z - c z^3 + o(z^3)$$ with $c=-1.23...$ This translates into a differential equation for the error (for the second iterate): $\Delta(n) = g^n(a+z_0)-a$: $$\frac{d\Delta}{dn} = -c\Delta^3 + ..$$ or solving by truncating to the leading term $$\Delta(n) = \Delta(0) \frac{1}{\sqrt{1+ c n^2 \Delta(0)^2}} \asymp \frac{1}{\sqrt{c}\;n}$$ but a rigorous bound for this error term is not simple.