# Explosive number(s) from $x_{n+1}=\left(1+\frac{1}{x_n}\right)^n$

I came accross a curiosity I don't fully understand. In an exam, the sequence $$\left(x_n\right)$$ is introduced such as for all $$n \geq 1$$ $$x_{n+1}=\left(1+\frac{1}{x_n}\right)^n$$ with $$x_1 = \alpha \in \left]0;+\infty\right[$$. It is argued that there exists one number $$\alpha^{\ast}$$, and only one, such that the sequence $$\left(x_n\right)$$ with $$x_1 = \alpha^{\ast}$$ diverges towards $$+\infty$$. A proof is then proposed where it is shown that $$\alpha^{\ast} \approx 1,1874$$. This result triggers my curiosity.

I've written a small python script, and I've observed that as soon as $$x_{N}$$ is "big", then $$x_{N+1} \approx 1$$. I guess all the mystery is there : how to have $$x_{N}$$ 'big' without having $$x_{N+1} \approx 1$$. For $$x_{N}$$ big enough, we can write

$$x_{N+1} = e^{N\ln\left(1+\frac{1}{x_N}\right)} = e^{N\left(\frac{1}{x_{N}} - \frac{1}{2x_{N}^2} + o\left(\frac{1}{x_{N}^2}\right)\right)}$$ So my guess is that $$N/x_{N}$$ needs to stay low. What I see for example is :

$$\begin{array} x_1 & 5 & 1.1874\newline x_2 & 1.20 & 1.84\newline x_3 & 3.36 & 2.38\newline x_4 & 2.18 & 2.86\newline x_5 & 4.52 & 3.31\newline x_6 & 2.72 & 3.74\newline x_7 & 6.55 & 4.14\newline x_8 & 2.70 & 4.54\newline x_9 & 12.4 & 4.92\newline x_{10} & 2.0 & 5.29\newline \end{array}$$

A pattern clearly emerges with $$\alpha = 1.1874$$, where odd and even indexes find a kind of balance. But after some more iterations, the balance is broken : we find $$x_{28} = 3.2 \cdot 10^{8}$$ and $$x_{29} \approx 1.0$$. Does it mean that it is not possible for a program to reproduce this pattern and just preserve it until a given rank ? Is there an equation that could be deduced for $$\alpha^{\ast}$$ ?

• Do you have a link to this proof and how they compute $a^*$? I'd imagine that the result is very unstable and requires a specific irrational number Apr 19, 2023 at 19:43
• Apr 19, 2023 at 19:43
• @MartinR Thanks ! I did not know that it existed ! Apr 19, 2023 at 19:50
• Apparently the precision of built-in floating point numbers is not enough for the late-20s terms. Consider using the "decimal" module if you want more precision. Apr 20, 2023 at 1:23