$\bf命题2：$设正项级数$\sum\limits_{n = 1}^\infty {{a_n}} $收敛，则存在发散到正无穷大的数列$\left\{ {{b_n}} \right\}$，使得级数$\sum\limits_{n = 1}^\infty {{a_n}{b_n}} $仍收敛
证明：令${r_n} = \sum\limits_{m = n}^\infty {{a_m}} $，则${r_n} \to 0\left( {n \to
\infty } \right)$，由于
\[\frac{{{a_n}}}{{\sqrt {{r_n}} }} = \frac{{{r_n} -
{r_{n + 1}}}}{{\sqrt {{r_n}} }} = \frac{{\left( {\sqrt {{r_n}} + \sqrt {{r_{n +
1}}} } \right)\left( {\sqrt {{r_n}} - \sqrt {{r_{n + 1}}} } \right)}}{{\sqrt
{{r_n}} }} \le 2\left( {\sqrt {{r_n}} - \sqrt {{r_{n + 1}}} }
\right)\]
而级数$\sum\limits_{n = 1}^\infty {2\left( {\sqrt {{r_n}} - \sqrt
{{r_{n + 1}}} } \right)}$收敛，所以由比较判别法知级数$\sum\limits_{n =
1}^\infty {\frac{{{a_n}}}{{\sqrt {{r_n}} }}} $收敛，取${b_n} = \frac{1}{{\sqrt
{{r_n}} }}$，则命题成立
原文地址：http://www.cnblogs.com/ly758241/p/3706457.html
