# Is it hard to find out for which $r$ the infinite power tower $r\uparrow r \uparrow r\uparrow \cdots$ converges?

It is well known that the infinite power tower $$r\uparrow r\uparrow r\uparrow\cdots$$ with $r>0$ converges if and only if $e^{-e}\le r\le e^{1/e}$.

I tried to prove it and I got stuck in the case $r\le 1/e$. I have to find out whether the fix-point-iteration $$x_1=r$$ $$x_{n+1}=r^{x_n}$$ converges to the solution of $x^r=r$ or not.

I assume that in the case $0<r<e^{-e}$, the iteration oscillates between two solutions of $r^{r^x}=x$. Is this right ? And if yes, how can I show that this happens ?

• If you read Latin, see Euler's De formulis exponentialibus replicatis: eulerarchive.maa.org//docs/originals/E489.pdf – punctured dusk Dec 8 '17 at 9:45
• What does $x\uparrow y$ mean again? Hyperexponentiation? – bof Dec 8 '17 at 10:52
• $x\uparrow y$ is simply $x^y$ – Peter Dec 8 '17 at 11:58
• Do you have access to the papers of Shell, Thron and so on? The articles are from the early 19xx 'th They study/prove even the range for convergence in the complex numbers. There is a literature database in the tetration-forum. I think that articles are accessible this way. If not put a further comment here. – Gottfried Helms Dec 9 '17 at 10:32