📝 题目
例 16 设两序列 $\left\{ {a}_{n}\right\} ,\left\{ {b}_{n}\right\}$ 满足关系式
$$ {a}_{n + 1} = {b}_{n} + q{a}_{n}\;\left( {0 < q < 1}\right) , $$
且 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b}$ 存在. 证明:
(1) $\left\{ {a}_{n}\right\}$ 有界； (2) $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}}$ 存在.
💡 答案与解析
证 (1) 因 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b,\exists M}$ ,使 $\left| {b}_{n}\right| \leq M\left( {n = 1,2,\cdots }\right)$ . 由关系式得
$$ \left| {a}_{2}\right| \leq M + q\left| {a}_{1}\right| , $$
$$ \left| {a}_{3}\right| \leq M + q\left| {a}_{2}\right| \leq M + {qM} + {q}^{2}\left| {a}_{1}\right| , $$
$\vdots$
$$ \left| {a}_{n}\right| \leq M\left( {1 + q + {q}^{2} + \cdots + {q}^{n - 2}}\right) + {q}^{n - 1}\left| {a}_{1}\right| $$
$$ \leq \frac{M}{1 - q} + \left| {a}_{1}\right| , $$
由数学归纳法即可看出式子成立.
(2) $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n + 1} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} + q\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \frac{b}{1 - q}}$ ,同理
$\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \frac{b}{1 - q} \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n}}$ 存在.