📝 题目
例 2 判别下列级数的收敛性:
(1) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{2}^{n - {\left( -1\right) }^{n}}}$ (2) $\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{a}^{n}}{1 + {a}^{2n}}\left( {a > 0}\right)$ .
💡 答案与解析
解 (1) $\mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{\frac{1}{{2}^{n - {\left( -1\right) }^{n}}}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{2}^{1 - {\left( -1\right) }^{n}/n}} = \frac{1}{2} < 1$ ,由柯西判别法知此级数收敛. 本题不能应用达朗贝尔判别法, 因为
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{{2n} + 1}}{{a}_{2n}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{2}^{{2n} - 1}}{{2}^{{2n} + 2}} = \frac{1}{8} < 1, $$
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{2n}}{{a}^{{2n} - 1}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{2}^{2n}}{{2}^{{2n} - 1}} = 2 > 1, $$
所以 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{a}_{n + 1}}{{a}_{n}}}$ 不存在.
(2)当 $a = 1$ 时,级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{2}}$ 显然发散.
当 $0 < a < 1$ 时,由
$$ \frac{{a}^{n}}{2} < \frac{{a}^{n}}{1 + {a}^{2n}} < {a}^{n} \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{\frac{{a}^{n}}{1 + {a}^{2n}}} = a < 1 \Rightarrow \text{ 级数收敛. } $$
当 $a > 1$ 时,因为
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{\frac{{a}^{n}}{1 + {a}^{2n}}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{\frac{{\left( 1/a\right) }^{n}}{1 + {\left( 1/a\right) }^{2n}}} = \frac{1}{a} < 1, $$
所以根据柯西判别法知级数收敛.