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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

∑k=mnaksk≥1sn∑k=mnak=1sn∑k=mn(sk?sk?1)=1sn(sn?sm?1)>1sn(sn?sm)

即对任意$n > m > 0$，固定$m$，有

∑k=mnaksk>sn?smsn

由$\lim \limits_{n \to \infty } {s_n} = + \infty $知，

limn→∞sn?smsn=limn→∞(1?smsn)=1

从而由极限的保号性知，存在$N > 0$，当$n > N$时，有

sn?smsn>12

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

sn>ε

特别地，取$\varepsilon = 1$，则${s_n} > 1$，从而可知当$n > N$时，有

ansnα>ansn

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

sn1?α?sn?11?α=(1?α)ξn?α(sn?sn?1)

从而可知

ansnα<ξn?α(sn?sn?1)=1α?1(sn?11?α?sn1?α)

于是

0<∑k=1nakskα≤a1s1α+1α?1(s11?α?sn1?α)<αα?1a11?α

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

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