📝 题目
例 5 求 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}$ .
💡 答案与解析
解
$$ 1 \leq \sqrt[n]{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}} \leq \sqrt[n]{n} $$
$$ \overset{\mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{n} = 1}{ = }\mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}} = 1. $$