Let $\alpha>0$ be a real number. Define the sequence $\alpha_{0}=\alpha, \ldots, \alpha_{n+1}=\alpha^{\alpha_{n}}, \ldots$. When does $\lim _{n \rightarrow \infty} \alpha_{n}$ exist?
Let's denote the function $a^{x}$ as $f(x)$. The sequence we obtain from the substitution

$$
\alpha_{n+1}=f\left(\alpha_{n}\right)
$$

can be geometrically visualized by also plotting the function $g(x)=x$ alongside the graph of $f(x)$. Using this, the ordinate of $\alpha_{n+1}$ can be converted into an abscissa, and we can find the value that $f$ takes at this point. Thus, starting from the point on the $x$-axis with abscissa $\alpha$, we "go up" to the point $(\alpha, f(\alpha))$, then move horizontally to the point $(\alpha_{1}, \alpha_{1})$, and finally move vertically to the point $\left(\alpha_{1}, f\left(\alpha_{1}\right)\right)$, continuing our path.

![](https://cdn.mathpix.com/cropped/2024_05_02_1c5b04876d1724fb8789g-1.jpg?height=1017&width=1027&top_left_y=527&top_left_x=538)

If $\alpha=1$, the value of $\alpha_{n}$ is also 1, and the terms of the sequence are equal. If $\alpha>1$, $f(x)$ is monotonically increasing, so the sequence is also increasing. Its fate is determined by whether $f(x)$ intersects $g(x)$ or not. If it does, the sequence "bumps into" this root, cannot step over it, and gradually approaches it. Since $f(0)=1>g(0)$, denoting the smallest positive root of the equation

$$
f(x)=g(x)
$$

by $\beta$, it is clear that $0<g^{\prime}(x)=1$; if $x>0$, the difference $f-g$ is monotonically increasing. If $\alpha=\alpha_{\max }$ such that the curves of $f$ and $g$ are tangent to each other, then $\alpha_{\max }$ is also a root of the equation

$$
f^{\prime}(x)=g^{\prime}(x)
$$

(We accept this based on geometric intuition, but it is not difficult to prove the correctness of this statement.) From the system of equations (1) and (2), we obtain that

$$
\alpha_{\max }=e^{1 / e}
$$

The sequence $\alpha_{n}$ is also monotonically increasing for $\alpha>\alpha_{\max }$, but it cannot be bounded, since otherwise it would be convergent, and (1) would have a root. Thus, in this case, it must be that

$$
\lim _{n \rightarrow \infty} \alpha_{n}=\infty
$$

If $0<f\left(\alpha_{1}\right)=\alpha_{2}>f\left(\alpha_{0}\right)=\alpha_{1}>f(1)=\alpha_{0}
$$

Continuing in this manner, we find that $0<\alpha_{n}<1$, and the sequence $\alpha_{2 k}$ is monotonically increasing, while $\alpha_{2 k+1}$ is monotonically decreasing. Thus, the sequence $\alpha_{n}$ is also convergent, and its limit is the only root of (1). Therefore, we find that if $0<\alpha \leq e^{1 / e}(=1.4446678)$, the sequence $\alpha_{n}$ is convergent, and its limit is the smaller positive root of (1). Otherwise, $\alpha_{n}$ tends to infinity.

Tar József (Eger, Gárdonyi G. Gymnasium)