📝 题目
例 6 设 $\alpha < 1$ ,求证: $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {{\left( n + 1\right) }^{\alpha } - {n}^{\alpha }}\right\rbrack = 0$ .
💡 答案与解析
证 当 $\alpha \leq 0$ 时,结论显然成立. 下设 $0 < \alpha < 1$ . 因为
$$ 0 < {\left( n + 1\right) }^{a} - {n}^{a} = {n}^{a}\left\lbrack {{\left( 1 + \frac{1}{n}\right) }^{a} - 1}\right\rbrack $$
$$ < {n}^{\alpha }\left\lbrack {{\left( 1 + \frac{1}{n}\right) }^{1} - 1}\right\rbrack = \frac{1}{{n}^{1 - \alpha }}, $$
又 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{n}^{1 - a} = 0}$ ,故有 $\mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {{\left( n + 1\right) }^{a} - {n}^{a}}\right\rbrack = 0$ .