Question 141
Given an integer $m \ge 2$, and two real numbers $a, b$ with $a \gt 0$ and $b \ne 0$, the sequence $\{x_n\}$ is such that $x_1 = b$ and $x_{n+1} = a x_n^m + b$, $n = 1, 2, \dots$. Prove that:
(1) When $b \lt 0$ and $m$ is even, the sequence $\{x_n\}$ is bounded if and only if $a b^{m-1} \ge -2$;
(2) When $b \lt 0$ and $m$ is odd, or when $b \gt 0$, the sequence $\{x_n\}$ is bounded if and only if $a b^{m-1} \le \frac{(m-1)^{m-1}}{m^m}$.
(1) When $b \lt 0$ and $m$ is even, the sequence $\{x_n\}$ is bounded if and only if $a b^{m-1} \ge -2$;
(2) When $b \lt 0$ and $m$ is odd, or when $b \gt 0$, the sequence $\{x_n\}$ is bounded if and only if $a b^{m-1} \le \frac{(m-1)^{m-1}}{m^m}$.