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I came across this problem that I would like to ask you about:

For which values $a>0$ does there $\exists$ a limit of the sequence $$a, a^{a},a^{a^{a}}, a^{a^{a^{a}}}...$$

Well this looks like a recursive sequence. If $a=exp(\lambda)$ then $z_{0}=0, z_{n+1}=\lambda exp(z_{n}).$

and I guess for some $\lambda$ $\exists$ limit. I'm just not sure how to show for which.

After some trying out my idea was to check for the border case $$a> e^{1/e} $$

and $$a \leq e^{1/e}$$

but I guess both diverge. Any idea or input is greatly appreciated! Thanks

Anixx
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uhoenig
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1 Answers1

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The example to bear in mind is $a = \sqrt 2,$ where the limit is $2.$ The point is that for some $1 < z < 2,$ we get $\sqrt2 < \left( \sqrt 2 \right)^z < 2,$ so the tower of exponential expressions gives an increasing sequence with an upper bound, therefore a limit. The limit, which I will call $Z,$ is no larger than $2$ and satisfies $ \left( \sqrt 2 \right)^Z = 2,$ so $Z=2.$ Note $\sqrt 2 \approx 1.41421.$

Similar for any $1 < a < e^{1/e} \approx 1.44466786.$ We take lower case $$ z_1 = a, \; z_2 = a^a = a^{z_1}, \; z_3 = a^{\left( a^a \right)} = a^{z_2}, \ldots, z_{n+1} = a^{z_n}. $$

The function $g(x) = \frac{\log x}{x}$ is strictly increasing on the interval $1 < x < e,$ with supremum of $1/e$ at the right endpoint. So, as $0 < \log a < 1/e$ we find there is a single value, I will call it $Z,$ with $1 < Z < e$ and $$ \frac{\log Z}{Z} = \log a. $$ This means $a^Z = Z.$

Now, take any $z$ with $1 < a \leq z <Z.$ We get $$ \frac{\log z}{z} < \frac{\log Z}{Z} = \log a. $$ So first we get $z < a^z.$ But from the elementary $z < Z$ we get $a^z < a^Z = Z.$ So we have $$ z < a^z < Z. $$

Returning to our numbered sequence $z_1, z_2, z_3, \ldots$ we find that the sequence is increasing with an upper bound of $Z.$ So the sequence has a limit.

By a continuity argument the limit satisfies $a^Z = Z,$ so the actual limit is $Z.$ Let me be a bit more specific on this point. The sequence has a limit, call it $W.$ Assume that $W < Z.$ The first thing the assumption tells us is that $a^W > W.$ Now, we know the sequence increases with $z_n \rightarrow W.$ For each $z_n,$ we also need $z_{n+1} = a^{z_n} \leq W.$ But we can get arbitrarily close to $W,$ indeed we can get all elements with $n \geq N$ between $W - \delta$ and $W$ for any $\delta > 0$ we like. So, what we actually need is $$ a^{W - \delta} \leq W, \; \; \forall \delta > 0. $$ And so, by continuity of $f(x) = a^x,$ we arrive at $$ a^W \leq W. $$ But this contradicts the assumption that $W < Z.$ So the actual limit is $Z.$

POSTSCRIPT: the bad news for making attractive problems with this is that $Z$ is in a certain price range, for example we cannot have $Z$ equal to three. Still, if we pick some $Z$ that we like, we can then solve for $a = Z^{1/Z}.$ For example, take $$ a = \left( \frac{5}{2} \right)^{\left( \frac{2}{5} \right)} \approx 1.4426999 $$ and the limit is $Z=\frac{5}{2}.$

Will Jagy
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