📝 题目
例 1 求级数 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{4{n}^{2} - 1}}$ 的和.
💡 答案与解析
解 因为
$$ {S}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}\frac{1}{\left( {{2k} - 1}\right) \left( {{2k} + 1}\right) } $$
$$ = \frac{1}{2}\mathop{\sum }\limits_{{k = 1}}^{n}\left\lbrack {\frac{1}{{2k} - 1} - \frac{1}{{2k} + 1}}\right\rbrack $$
$$ = \frac{1}{2}\left( {1 - \frac{1}{{2n} + 1}}\right) , $$
故 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} = 1/2}$ . 所以级数收敛,其和为 $1/2$ .