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Provided that $x$ is positive, when does the following converge? $$x^{x^{x^{x^{...}}}}$$

I just had a talk with my friend who is taking a calculus class and he is dealing a lot with divergence and convergence stuff. His TA said that if anyone could answer this and write a proof, he or she doesn't have to come to the recitation class anymore. This drives me a lot.

So here is my work, but I don't really know if this is correct. Could anyone shed some light on this?

I haven't done any real analysis before so I'm trying to make use of whatever I can use, and my plan is to find the critical points of $y'$ and look at the graph of $y=f(x)$. First let $f(x)=x^{x^{x^{x^{...}}}}=y=x^y$, then I have $$ \ln{y}=y\ln{x} $$ $$ \frac{d(\frac{\ln{y}}{y})}{dx} = \frac{d\ln{x}}{dx} $$ $$ \frac{1-\ln{y}}{y^2} \frac{dy}{dx} = \frac{1}{x}$$ $$ y'= \frac{f^2(x)}{x(1-\ln{f(x)})} $$ Now I noticed that the critical points are $x=0$ and $f(x)=0$ or $f(x)=e$. Since $f(x)=x^{x^{x^{x^{...}}}}$ is an increasing function, the graph of $y=f(x)$ would go up vertically at some point as $x$ increases, and that is when the function diverges. From these 3 critical points, $f(x)=e$ seems to make most sense. By plugging $f(x)=e$ back into $y=x^y$, it's not hard to see that $f(x)=e$ occurs at $x=e^{\frac{1}{e}}$. So the function converges when $x\in (0, e^{\frac{1}{e}})$

Edit: I just noticed something strange in my proof: $$ y'= \frac{f^2(x)}{x(1-\ln{f(x)})} $$ When $f(x)>e$, the denominator on the right hand side is negative. But since $f(x)$ is an increasing function, therefore, shouldn't $y'$ be positive? Now it seems to deep for me to perceive...

IgNite
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    Why should $f$ be differentiable in the first place? – Travis Willse Apr 14 '16 at 14:15
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    You may wish to see this question. – Eff Apr 14 '16 at 14:15
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    The upper bound is correct. For the lower, try $e^{-e}$ – Deepak Apr 14 '16 at 14:15
  • I always like to phrase the question as:

    Solve $x^{x^{x^{x^{...}}}} = 4$.

    Now solve $x^{x^{x^{x^{...}}}} = 2$.

     – Julian Braun Apr 14 '16 at 14:20
  • Why $0<x<e^{-e}$ doesn't make the function converge? I ask this because it seems to me that this should converge just by inspection. – IgNite Apr 14 '16 at 14:23
  • Inspection. What a great (but phony) reason. – GEdgar Apr 14 '16 at 14:24
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    It does indeed even converge for $(0,e^{\frac{1}{e}}]$. Your proof is not very rigorous anyway (differentiability,...). Better proof: define $x_{n+1}=x^{x_n}$, then proof by induction that $x_n \leq e$ for $x \leq e^{\frac{1}{e}} $ – Julian Braun Apr 14 '16 at 14:30
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    @JulianBraun It does not converge there. The sequence should be $a_{n+1} = a_{n} ^{x}$ with $a_{0} = x$. With $x \in (0, e^{-e})$ the series alternates, $a_{2n}$ and $a_{2n+1}$ converge to different values. – kalhartt Apr 14 '16 at 15:15
  • According to Wikipedia, it converges for $e^{-e}\le x\le e^{1/e}$. This was proven by Euler. – Akiva Weinberger Apr 14 '16 at 15:16
  • @kalhartt you really need to define it like I already did above. $(x^x)^x = x^{(x \cdot x)}$! But Akiva is right with the convergence interval. I actually meant $[1,e^{\frac{1}{e}}]$, this is where the $x_n$ are just monotone increasing and bounded. – Julian Braun Apr 14 '16 at 21:15
  • @JulianBraun Yes, I mistakenly swapped $a_n$ and $x$ on the rhs above. My point is that the lower bound is not 0 or 1 but $e^{-e}$ inclusive and that boundedness is not enough to show convergence here. Also using $x_n$ as a symbol for the finite tetration can be confusing when $x$ is already used as a single value. – kalhartt Apr 15 '16 at 15:10

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