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This question arose from here. I am interested to find a nice proof about the convergence of $${^n}a=\underbrace{a^{a^{\ .^{\ .^{\ .^a}}}}}_{n\ \text{times}}.$$ I find with google a necessary and sufficient condition to have the convergence is $\frac{1}{e^e} \leq a \leq e^{1/e}$ but the part for $a\le1$ need some ugly work.

Does anyone have an elegant/slick proof ?

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  • $\begingroup$ For finite $n$ I imagine it'll always converge, do you mean infinite $n$? $\endgroup$ – frogeyedpeas Aug 7 '14 at 18:27
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    $\begingroup$ @frogeyedpeas Sure, I mean $\lim\limits_{n\to\infty}{^n}a$. $\endgroup$ – Free X Aug 7 '14 at 18:30
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Does anyone have an elegant/slick proof ?

Like in the answer given to the link you quote by Kiril, the fastest proof is given by fixed point iteration. The function you iterate, is:

$$g_c(x)=c^x$$

The fixed points of $g_c(x)$ are given in terms of the Lambert $W$ function as:

$$x_0=\frac{W(-\ln(c))}{-\ln(c)}$$

The fixed point condition (for attraction hence convergence) is therefore:

$$|g'(x_0)|\le 1\Rightarrow$$ $$|-W(-\ln(c))|\le 1 (*)$$

Now consider the function:

$$m(x)=x\cdot\exp(x)$$

$$(*)\Rightarrow W(-\ln(c))\in [-1,1]\Rightarrow$$ $$m(W(-\ln(c)))\in m([-1,1])\Rightarrow$$ $$m(W(-\ln(c)))\in [-e^{-1},e]\Rightarrow$$ $$-\ln(c)\in[-e^{-1},e]\Rightarrow$$ $$c\in[e^{-e},e^{1/e}]$$

Fixed point iteration, therefore, takes care of all cases where $c\in(e^{-e},e^{1/e})$ (open interval). The two end points have to be checked separately, because the fixed point condition is inconclusive there, being possibly $\pm1$.

The upper end point ($e^{1/e}$) is easy using standard Calculus (Consider the sequence $a_n=g_{e^{1/e}}^{(n)}(1)$ and prove that it converges to $e$, etc.). The lower end point ($e^{-e}$), is a bit tricky, but can also be done using Calculus, and you are done.

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Firstly if $$a \le -1$$ then we are guaranteed $$1 \le |a^a| \le 0$$ so if the value converges it is definitely bounded

For $$-1 < a \le 0$$ we have that $$ |a^a| \rightarrow 1 $$ so again boundedness isn't an issue

For $$0 < a \le 1$$ the same reasoning holds to ensure boundedness

Now for convergence notice the following. If we assume the tower converges to a finite value, that is:

$$ a^{a^{a^{a^{\vdots}}}} = n$$

Then we can take logarithms on both sides and use the power to product rule to find

$$ a^{a^{a^{a^{\vdots}}}}\log(a) = \log(n)$$

In other words:

$$n \log(a) = \log(n)$$

Exponentiate both sides

$$ a^n = n$$

Yielding:

$$a = n^{\frac{1}{n}}$$

Now we can use calculus to find the maximum and minimal values that $n^{\frac{1}{n}}$ take on and this gives us the range of possible $a$ among the real numbers if we want $n$ to be among the real numbers.

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  • $\begingroup$ $n^{\frac {1}{n}}$ has doesn't have a positive minimum, then how would this explain the $e^{-e}$ part? $\endgroup$ – Dhvanit Apr 13 at 4:04

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