68

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

证明：$left( 1 ight)$当$alpha = 1$时，由正项级数$sumlimits_{n = 1}^infty{{a_n}} $发散知，$lim limits_{n o infty } {s_n} = + infty $且$left{ {{s_n}} ight}$严格递增，于是
egin{align*}
sumlimits_{k = m}^n {frac{{{a_k}}}{{{s_k}}}} &ge frac{1}{{{s_n}}}sumlimits_{k = m}^n {{a_k}} \&
= frac{1}{{{s_n}}}sumlimits_{k = m}^n {left( {{s_k} - {s_{k - 1}}} ight)} \&
= frac{1}{{{s_n}}}left( {{s_n} - {s_{m - 1}}} ight) > frac{1}{{{s_n}}}left( {{s_n} - {s_m}} ight)
end{align*}即对任意$n > m > 0$，固定$m$，有[sumlimits_{k = m}^n {frac{{{a_k}}}{{{s_k}}}} > frac{{{s_n} - {s_m}}}{{{s_n}}}]
由$lim limits_{n o infty } {s_n} = + infty $知，
[mathop {lim }limits_{n o infty } frac{{{s_n} - {s_m}}}{{{s_n}}} = mathop {lim }limits_{n o infty } left( {1 - frac{{{s_m}}}{{{s_n}}}} ight) = 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$sumlimits_{k = m}^n {frac{{{a_k}}}{{{s_k}}}} > frac{1}{2}$，由$f{Cauchy收敛准则}$知，级数$sumlimits_{n = 1}^infty {frac{{{a_n}}}{{{s_n}}}} $发散

$left( 2 ight)$当$alpha < 1$时，由$lim limits_{n o 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$时，有
[frac{{{a_n}}}{{{s_n}^alpha }} > frac{{{a_n}}}{{{s_n}}}]
而$sumlimits_{n = 1}^infty {frac{{{a_n}}}{{{s_n}}}}$发散，由$f比较判别法$知，$sumlimits_{n = 1}^infty {frac{{{a_n}}}{{{s_n}^alpha }}} $发散

$left( 3 ight)$当$alpha > 1$时，设$fleft( x ight) = {x^{1 - alpha }}$，则由微分中值定理知，存在${xi _n} in left( {{s_{n -1}},{s_n}} ight)$，使得
[{s_n}^{1 - alpha } - {s_{n - 1}}^{1 - alpha } = left( {1 - alpha } ight){xi _n}^{ - alpha }left( {{s_n} - {s_{n - 1}}} ight)]
从而可知[frac{{{a_n}}}{{{s_n}^alpha }} < {xi _n}^{ - alpha }left( {{s_n} - {s_{n - 1}}} ight) = frac{1}{{alpha - 1}}left( {{s_{n - 1}}^{1 - alpha } - {s_n}^{1 - alpha }} ight)]于是[0 < sumlimits_{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 }} ight) < 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$sumlimits_{n = 1}^infty {frac{{{a_n}}}{{{s_n}^alpha }}} $收敛

$f注1：$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

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