Let $a > 0$. Show that the sequence defined by $$ x_0 = 1, \qquad x_{n+1} = a^{x_n} $$ converges for $a \leq e^{1/e}$.

Any help is appreciated, I don't even know where to start with this.

Edit to add clarifications:

The case $1 \leq a \leq e^{1/e}$ is solved below in the comments and answers. It is also easy to show that the sequence does not converge for $a > e^{1/e}$ (not part of the question but still worth mentioning).

The interesting case is the remaining one $0 < a < 1$. Here, I found that $x_{n-1} < x_n \Rightarrow x_n > x_{n+1}$ so that the sequence seems to be oscillating. Since $x_0 > x_1$, this implies that the subsequence $\{x_{2n}\}$ is decreasing while the subsequence $\{x_{2n+1}\}$ is increasing, and both converge since they are bounded. This is illustrated by the plot below of the first few terms of the sequence when $a = 0.1$. plot with a = 0.1

It remains to show that both subsequences have the same limit, which is where I have trouble.

  • $\begingroup$ The sequence is monotonic increasing for $1 < a \leq e^{1/e}$, it's left to find the upper bound. $\endgroup$ – Azlif Nov 9 '19 at 15:43

The convergence of the sequence is determined by the properties of the function $f(x)=a^x.$ Below is a sketch. A thorough analysis may be found here.

$1).\ $First suppose that $1<a<e^{1/e}.$

Then, $x_1>x_0$ and if $x_k\ge x_{k-1}$ for all $k\le n,$ then $\frac{x_{n+1}}{x_n}=a^{x_n-x_{n-1}}>1$ so $(x_n)$ is increasing.

Now, $x_0\le e$. Assume that $x_k\le e$ for all $k\le n$. Then, $x_{n+1}=a^{x_n}\le e^{x_n/e}\le e.$

So $(x_n)$ is increasing and bounded, so $x_n\to l\in [0,e]$.

$2).\ $ Now suppose that $e^{-e}<a<1.$ Then, as you pointed out, the even terms are decreasing and the odd terms are increasing. The sequence is easily seen to be bounded, and so $\limsup x_n=l$ and $\liminf x_n=m$ are real numbers. And because the even and odd subsequences are monotone, if $l\neq m,$ then we have $|x_n-x_{n-1}|>(l-m)/2$ if $n$ is large enough.

But now, note that the tower $a^{\scriptscriptstyle a^{\cdot^{a ^{\cdot^{a}}}}}$ converges (see for example this article)

$|x_n-x_{n-1}|=\left|a^{\scriptscriptstyle a^{\cdot^{a ^{\cdot^{x_1}}}}}-a^{\scriptscriptstyle a^{\cdot^{a ^{\cdot^{x_0}}}}}\right |=\left |a^{\scriptscriptstyle a^{\cdot^{a ^{\cdot^{a}}}}}-a^{\scriptscriptstyle a^{\cdot^{a ^{\cdot^{1}}}}}\right |\to 0$ as $n\to \infty$ which is a contradiction and so in fact $l=m.$

$3).\ $ if $0<a<e^{-1}$, the sequence diverges.

  • $\begingroup$ Sorry, why is it increasing when $a < 1$? $\endgroup$ – häxq Nov 18 '19 at 12:50
  • $\begingroup$ Yes, you are right. My calculation was wrong. Apologies. I will repost. $\endgroup$ – Matematleta Nov 18 '19 at 23:43

Notice that if your sequence has a limit $L$, then $a^L=L$. Using a bit of algebra, this is equivalent to saying that $-\ln a =(-L\ln a)e^{-L\ln a}$. Then, you may use Lambert's W function so that $-L\ln a=W(-\ln a)$, and so $$L=\frac{-W(-\ln a)}{\ln a}.$$

However, this $W$ function is defined (for real values) only when $x\geq -1/e$. Moreover, its argument, $-\ln a$, is a decreasing function with $-\ln(e^{1/e})=-1/e$, so we must have $a\leq e^{1/e}$.


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