📝 题目
证 对 $\forall n,k \in N$ ,注意到
$$ \mathop{\sum }\limits_{{n = 2}}^{k}\frac{{a}_{n}}{{n}^{2} + {m}^{2}} \leq {\int }_{1}^{k}\frac{{a}_{n}}{{n}^{2} + {x}^{2}}\mathrm{\;d}x $$
$$ = \frac{{a}_{n}}{n}\left( {\arctan \frac{k}{n} - \arctan \frac{1}{n}}\right) < \frac{\pi }{2} \cdot \frac{{a}_{n}}{n}, $$
于是有
$$ {c}_{n}\overset{\text{ 定义 }}{ = }\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}} = \frac{{a}_{n}}{{n}^{2} + 1} + \mathop{\sum }\limits_{{m = 2}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}} $$
$$ \leq \frac{1}{2}\left( {\pi + 1}\right) \frac{{a}_{n}}{n}, $$
于是 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}$ 收敛,即 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}}}$ 收敛.
💡 答案与解析
证 对 $\forall n,k \in N$ ,注意到
$$ \mathop{\sum }\limits_{{n = 2}}^{k}\frac{{a}_{n}}{{n}^{2} + {m}^{2}} \leq {\int }_{1}^{k}\frac{{a}_{n}}{{n}^{2} + {x}^{2}}\mathrm{\;d}x $$
$$ = \frac{{a}_{n}}{n}\left( {\arctan \frac{k}{n} - \arctan \frac{1}{n}}\right) < \frac{\pi }{2} \cdot \frac{{a}_{n}}{n}, $$
于是有
$$ {c}_{n}\overset{\text{ 定义 }}{ = }\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}} = \frac{{a}_{n}}{{n}^{2} + 1} + \mathop{\sum }\limits_{{m = 2}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}} $$
$$ \leq \frac{1}{2}\left( {\pi + 1}\right) \frac{{a}_{n}}{n}, $$
于是 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }{c}_{n}}$ 收敛,即 $\displaystyle{\mathop{\sum }\limits_{{n = 1}}^{\infty }\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{{a}_{n}}{{n}^{2} + {m}^{2}}}$ 收敛.