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I've been having fun with the problem of finding the values of $n$ for which the infinite power tower $$\sqrt{2}^{\sqrt{2}^{...^{\sqrt{2}^n}}}$$ Has a finite value. My final answer was that it converged to a finite number for $n\leq4$. I reasoned that whenever $\sqrt2$ was raised to a power between $1$ and $2$, the result would also be between $1$ and $2$, so if a value occurs anywhere along the height of the tower, the whole thing would end up being between $1$ and $2$ (converging to 2). However, now that I'm trying the same problem with a power tower of $\sqrt3$, I can't determine when it converges because $\sqrt3$ raised to a power between $1$ and $2$ can be greater than $2$. I suspect that it may not ever converge, but how do I prove this? Help?

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    $\begingroup$ That's because it diverges. $\endgroup$ Commented Apr 24, 2017 at 21:40
  • $\begingroup$ If it diverges, how do I prove it? $\endgroup$ Commented Apr 24, 2017 at 21:50
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    $\begingroup$ Wikipedia article on tetration discusses infinite heights in the section "Extension to infinite heights". It gives the range of convergence as $e^{-e} \leq x \leq e^{1/e}$. Since $\sqrt{3}$ is outside the range, it doesn't converge. $\endgroup$
    – Χpẘ
    Commented Apr 24, 2017 at 22:03

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If your sequence converges, the limit must necessarily be a solution to $$ (\sqrt 3)^L = L $$ But this equation has no real solution -- just plot $(\sqrt 3)^x-x$ and see that it is always positive.

($B^L=L$ has a solution for $L$ if and only if $0<B\le e^{1/e}\approx 1.44467$).

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  • $\begingroup$ Check the Wikipedia article I cited in comments to OP. Lower value of convergence range is $e^{-e}$ $\endgroup$
    – Χpẘ
    Commented Apr 24, 2017 at 22:05
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    $\begingroup$ @Χpẘ: Sure -- but here I was speaking narrowly about whether there is a fixpoint, not about whether that fixpoint is actually a limit. $\endgroup$ Commented Apr 24, 2017 at 22:07
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There isn't any need to put that much effort into finding the solution. We see that since it is a self repeating process of operations, we can give the infinite tower a value of $x$. We derive,

$x = \sqrt{3}^{x}$

$x^2 = 3^x$

After this step, the value of x can be approximated.

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  • $\begingroup$ Also, the answer to your $\sqrt{2}$ tower in this similar method is $4$ and not $2$. The limit can also be checked in the calculator. It neatly approaches $4$. $\endgroup$
    – Haran
    Commented May 17, 2017 at 10:24

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