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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...

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    $\begingroup$ Why should $f$ be differentiable in the first place? $\endgroup$
    – Travis Willse
    Apr 14, 2016 at 14:15
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    $\begingroup$ You may wish to see this question. $\endgroup$
    – Eff
    Apr 14, 2016 at 14:15
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    $\begingroup$ The upper bound is correct. For the lower, try $e^{-e}$ $\endgroup$
    – Deepak
    Apr 14, 2016 at 14:15
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    $\begingroup$ 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}} $ $\endgroup$
    – Julian Braun
    Apr 14, 2016 at 14:30
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    $\begingroup$ @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. $\endgroup$
    – kalhartt
    Apr 14, 2016 at 15:15

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