套娃式题目怎么做？先“套”几个看看

一、题目

已知， $1 < a ⩽ e^{\frac{1}{e}}$ , $x_{1} = a, x_{2} = a^{x_{1}}, \dots, x_{n} = a^{x_{n - 1}}, \dots$ , 则数列 ${x_{n}}$ 增减性如何？有极限吗？

难度评级：

$a = e^{\frac{1}{e}} \Rightarrow x_{1} = e^{\frac{1}{e}} \Rightarrow x_{2} = {(e^{\frac{1}{e}})}^{\frac{1}{e}} \Rightarrow$

$x_{2} = e^{\frac{1}{e}} \cdot e^{\frac{1}{e}} \Rightarrow x_{3} = e^{\frac{1}{e}} \cdot e^{\frac{1}{e}} \cdot e^{\frac{1}{e}}$

$e^{\frac{1}{e}} > 1 \Rightarrow e^{\frac{1}{e} \cdot e^{\frac{1}{e}}} > e^{\frac{1}{e}} \Rightarrow$

$e^{\frac{1}{e} \cdot e^{\frac{1}{e}} \cdot e^{\frac{1}{e}}} > e^{\frac{1}{e} \cdot e^{\frac{1}{e}}}$

$x_{n} = e^{\frac{1}{e} \cdot e^{\frac{1}{e^{n - 1}}}} < e$

$\frac{1}{e} \cdot e^{\frac{1}{e^{n - 1}}} < 1 \Rightarrow$

数列 ${x_{n}}$ 单调递增有极限。

$a > 1 \Rightarrow x_{2} = a^{x_{1}} > a = x_{1} \Rightarrow$

$x_{2} > 1 \Rightarrow$

$x_{3} = a^{x_{2}} > x_{2} > 1 \Rightarrow$

$x_{n} > x_{n - 1}$

$a \leq e^{\frac{1}{e}} \Rightarrow x_{1} = a \leq e \Rightarrow$

若设 $x_{n} < e$ , 则：

$x_{n + 1} = a^{x_{n}} ⩽ e$

因此，数列 ${x_{n}}$ 单调递增有极限。