📝 题目
例 19 设 ${a}_{n} > 0$ ,且 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{n + 1}}{{a}_{n}} = a}$ ,求证: $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{{a}_{n}} = a}$ .
💡 答案与解析
证 改写
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{{a}_{n}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{\frac{{a}_{n}}{{a}_{n - 1}} \cdot \frac{{a}_{n - 1}}{{a}_{n - 2}} \cdot \cdots \cdot \frac{{a}_{2}}{{a}_{1}} \cdot {a}_{1}} $$
$$ = \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{{a}_{1}} \cdot \sqrt[n]{\frac{{a}_{n}}{{a}_{n - 1}} \cdot \frac{{a}_{n - 1}}{{a}_{n - 2}} \cdot \cdots \cdot \frac{{a}_{2}}{{a}_{1}}} $$
$$ = \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{{a}_{1}} \cdot {\left\lbrack {\left( \frac{{a}_{n}}{{a}_{n - 1}} \cdot \frac{{a}_{n - 1}}{{a}_{n - 2}} \cdot \cdots \cdot \frac{{a}_{2}}{{a}_{1}}\right) }^{\frac{1}{n - 1}}\right\rbrack }^{\frac{n - 1}{n}} $$
用