# Convergence of the sequence $x_{n+1} = a^{x_n}$ [duplicate]

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.

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 $$a = 0.1$$. It remains to show that both subsequences have the same limit, which is where I have trouble.

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

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} 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, the sequence diverges.

• Sorry, why is it increasing when $a < 1$? – häxq Nov 18 at 12:50
• Yes, you are right. My calculation was wrong. Apologies. I will repost. – Matematleta Nov 18 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 $$a\leq e^{1/e}$$.