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$\bf命题1：$设正项级数$\sum\limits_{n = 1}^\infty {{a_n}} $发散，且${s_n} = \sum\limits_{k = 1}^n {{a_k}} $，试讨论级数$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{{s_n}^\alpha }}} \ $的敛散性

证明：$\left( 1 \right)$当$\alpha = 1$时，由正项级数$\sum\limits_{n = 1}^\infty{{a_n}} $发散知，$\lim \limits_{n \to \infty } {s_n} = + \infty $且$\left\{ {{s_n}} \right\}$严格递增，于是
\begin{align*}
\sum\limits_{k = m}^n {\frac{{{a_k}}}{{{s_k}}}} &\ge \frac{1}{{{s_n}}}\sum\limits_{k = m}^n {{a_k}} \\&
= \frac{1}{{{s_n}}}\sum\limits_{k = m}^n {\left( {{s_k} - {s_{k - 1}}} \right)} \\&
= \frac{1}{{{s_n}}}\left( {{s_n} - {s_{m - 1}}} \right) > \frac{1}{{{s_n}}}\left( {{s_n} - {s_m}} \right)
\end{align*}即对任意$n > m > 0$，固定$m$，有\[\sum\limits_{k = m}^n {\frac{{{a_k}}}{{{s_k}}}} > \frac{{{s_n} - {s_m}}}{{{s_n}}}\]
由$\lim \limits_{n \to \infty } {s_n} = + \infty $知，
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{s_n} - {s_m}}}{{{s_n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{{{s_m}}}{{{s_n}}}} \right) = 1\]

从而由极限的保号性知，存在$N > 0$，当$n > N$时，有
\[\frac{{{s_n} - {s_m}}}{{{s_n}}} > \frac{1}{2}\]
即$\sum\limits_{k = m}^n {\frac{{{a_k}}}{{{s_k}}}} > \frac{1}{2}$，由$\bf{Cauchy收敛准则}$知，级数$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{{s_n}}}} $发散

$\left( 2 \right)$当$\alpha < 1$时，由$\lim \limits_{n \to \infty } {s_n} = + \infty $知，对任意$\varepsilon > 0$，存在$N > 0$，当$n > N$时，有\[{s_n} > \varepsilon \]
特别地，取$\varepsilon = 1$，则${s_n} > 1$，从而可知当$n > N$时，有
\[\frac{{{a_n}}}{{{s_n}^\alpha }} > \frac{{{a_n}}}{{{s_n}}}\]
而$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{{s_n}}}}$发散，由$\bf比较判别法$知，$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{{s_n}^\alpha }}} $发散

$\left( 3\right)$当$\alpha > 1$时，设$f\left( x \right) = {x^{1 - \alpha }}$，则由微分中值定理知，存在${\xi _n} \in \left( {{s_{n -1}},{s_n}} \right)$，使得
\[{s_n}^{1 - \alpha } - {s_{n - 1}}^{1 - \alpha } = \left( {1 - \alpha } \right){\xi _n}^{ - \alpha }\left( {{s_n} - {s_{n - 1}}} \right)\]
从而可知\[\frac{{{a_n}}}{{{s_n}^\alpha }} < {\xi _n}^{ - \alpha }\left( {{s_n} - {s_{n - 1}}} \right) = \frac{1}{{\alpha - 1}}\left( {{s_{n - 1}}^{1 - \alpha } - {s_n}^{1 - \alpha }} \right)\]于是\[0 < \sum\limits_{k = 1}^n {\frac{{{a_k}}}{{{s_k}^\alpha }}} \le \frac{{{a_1}}}{{{s_1}^\alpha }} + \frac{1}{{\alpha - 1}}\left( {{s_1}^{1 - \alpha } - {s_n}^{1 - \alpha }} \right) < \frac{\alpha }{{\alpha - 1}}{a_1}^{1 - \alpha }\]
从而由正项级数收敛的基本定理知，级数$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{{s_n}^\alpha }}} $收敛

$\bf注1：$正项级数收敛的基本定理：正项级数收敛当且仅当其部分和数列有界

$\bf注2：$我们可得到下面命题：设正项级数$\sum\limits_{n = 1}^\infty {{a_n}} $发散，则存在收敛于$0$的正项数列$\left\{ {{b_n}} \right\}$，使得级数$\sum\limits_{n = 1}^\infty {{a_n}{b_n}} $仍发散