📝 题目
例 12 设 $\displaystyle{\lim {x}_{n} = a}$ . 求证
$$ \lim \frac{{x}_{1} + 2{x}_{2} + \cdots + n{x}_{n}}{{n}^{2}} = \frac{a}{2}. $$
💡 答案与解析
证明 我们有
$$ {x}_{n} = a + {\alpha }_{n}, $$
这里 $\left\{ {\alpha }_{n}\right\}$ 是无穷小序列. 于是
$$ \frac{{x}_{1} + 2{x}_{2} + \cdots + n{x}_{n}}{{n}^{2}} $$
$$ = \frac{\left( {a + {\alpha }_{1}}\right) + 2\left( {a + {\alpha }_{2}}\right) + \cdots + n\left( {a + {\alpha }_{n}}\right) }{{n}^{2}} $$
$$ = \frac{n + 1}{2n}a + \frac{\frac{1}{n}{\alpha }_{1} + \frac{2}{n}{\alpha }_{2} + \cdots + \frac{n}{n}{\alpha }_{n}}{n}. $$
因为
$$ \left| \frac{\frac{1}{n}{\alpha }_{1} + \frac{2}{n}{\alpha }_{2} + \cdots + \frac{n}{n}{\alpha }_{n}}{n}\right| \leq \frac{\left| {\alpha }_{1}\right| + \left| {\alpha }_{2}\right| + \cdots + \left| {\alpha }_{n}\right| }{n}, $$
所以
$$ \lim \frac{{x}_{1} + 2{x}_{2} + \cdots + n{x}_{n}}{{n}^{2}} $$
$$ = \lim \frac{n + 1}{2n}a + \lim \frac{\frac{1}{n}{\alpha }_{1} + \frac{2}{n}{\alpha }_{2} + \cdots + \frac{n}{n}{\alpha }_{n}}{n} $$
$$ = \frac{a}{2} + 0 = \frac{a}{2}. $$