📝 题目
例 11 设 $\displaystyle{\lim {x}_{n} = a,\lim {y}_{n} = b}$ . 若记
$$ {u}_{n} = \frac{{x}_{1}{y}_{n} + {x}_{2}{y}_{n - 1} + \cdots + {x}_{n}{y}_{1}}{n},\;n = 1,2,\cdots , $$
则有
$$ \lim {u}_{n} = {ab}. $$
💡 答案与解析
证明 我们有
$$ {x}_{n} = a + {\alpha }_{n},\;{y}_{n} = b + {\beta }_{n}, $$
这里的 $\left\{ {\alpha }_{n}\right\}$ 和 $\left\{ {\beta }_{n}\right\}$ 是无穷小序列. 于是
$$ {u}_{n} = \frac{\left( {a + {\alpha }_{1}}\right) \left( {b + {\beta }_{n}}\right) + \cdots + \left( {a + {\alpha }_{n}}\right) \left( {b + {\beta }_{1}}\right) }{n} $$
$$ = {ab} + \frac{{\alpha }_{1} + \cdots + {\alpha }_{n}}{n}b + a\frac{{\beta }_{1} + \cdots + {\beta }_{n}}{n} $$
$$ + \frac{{\alpha }_{1}{\beta }_{n} + \cdots + {\alpha }_{n}{\beta }_{1}}{n}\text{ . } $$
无穷小序列也是有界序列, 可设
$$ \left| {\beta }_{n}\right| \leq L,\;\forall n \in \mathbb{N}. $$
因为
$$ \left| \frac{{\alpha }_{1}{\beta }_{n} + \cdots + {\alpha }_{n}{\beta }_{1}}{n}\right| \leq \frac{\left| {\alpha }_{1}\right| + \cdots + \left| {\alpha }_{n}\right| }{n} \cdot L, $$
所以
$$ \left\{ \frac{{\alpha }_{1}{\beta }_{n} + \cdots + {\alpha }_{n}{\beta }_{1}}{n}\right\} $$
是无穷小序列. 又因为
$$ \left\{ {\frac{{\alpha }_{1} + \cdots + {\alpha }_{n}}{n}b}\right\} ,\;\left\{ {a\frac{{\beta }_{1} + \cdots + {\beta }_{n}}{n}}\right\} $$
也都是无穷小序列, 所以
$$ \lim {u}_{n} = {ab}. $$