The iteration $$x_1=r$$ $$x_{n+1}=r^{x_n}$$ with $r=e^{e^{-1}}$ tends to $e$.

What is the smallest index $n$ such that $|x_n-e|<\epsilon$ ?

For small $\epsilon$, it seems that the smallest such index is approximately $\large \color\red{\frac{2e}{\epsilon}}$. Is this true, and if yes, how can it be proven ?

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    $\begingroup$ Don't know if this is useful, but $\ln x_n - \ln e = \ln\ln x_{n+1}$ $\endgroup$ – rogerl Dec 7 '17 at 16:36
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    $\begingroup$ For $r=e^{e^{-1}}$ We have $x_{n+1} = \exp(x_n e^{-1}) = f(x_n)$ with $f(x) =\exp(e^{-1}x) \implies f'(x) =e^{-1}\exp(e^{-1}x)$ We have, $$|x_{n+1} -e| = |f(x_n)-f(e)|\\\le e^{-1}|\int_e^{x_n} \exp(e^{-1}x)dx| \\\le |x_n-e|$$ may this help $\endgroup$ – Guy Fsone Dec 7 '17 at 16:45

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