📝 题目
例 1 考察级数
$$ \mathop{\sum }\limits_{{n = 1}}^{{+\infty }}\frac{1}{{n}^{2}} $$
的敛散性.
💡 答案与解析
解 因为
$$ \mathop{\sum }\limits_{{n = 1}}^{N}\frac{1}{{n}^{2}} \leq 1 + \mathop{\sum }\limits_{{n = 2}}^{N}\frac{1}{n\left( {n - 1}\right) } = 1 + \mathop{\sum }\limits_{{n = 2}}^{N}\left( {\frac{1}{n - 1} - \frac{1}{n}}\right) $$
$$ = 2 - \frac{1}{N} \leq 2,\;\forall N \in \mathbb{N}, $$
所以级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{{+\infty }}\frac{1}{{n}^{2}}}$ 收敛.