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    $f命题1:$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

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  • 原文地址:https://www.cnblogs.com/ly758241/p/3706448.html
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