Abel Transform

对数列 $\left\{a_n\right\},\left\{b_n\right\}$,记$S_k=\sum _{i=1}^k{a}_i,k=1,2,\cdots ,n,S_0=0,$则有$\sum _{k=1}^n{a}_k{b}_k=S_n{b}_n+\sum _{k=1}^{n-1}{S}_k(b_k-b_{k+1})$.

上式称为阿贝尔变换,或分部求和公式(类似于分部积分),它可用于证明无穷级数的阿贝尔判别法. 证明:由$a_k=S_k-S_{k-1}$得$$\begin{array}{l}\sum _{k=1}^n{a}_k{b}_k=\sum _{k=1}^n(S_k-S_{k-1})b_k\\ =\sum _{k=1}^n{S}_k{b}_k-\sum _{k=1}^n{S}_{k-1}{b}_k\\ =\sum _{k=1}^n{S}_k{b}_k-\sum _{k=1}^{n-1}{S}_k{b}_{k+1}\\ =S_n{b}_n+\sum _{k=1}^{n-1}{S}_k(b_k-b_{k+1})\end{array}$$应用阿贝尔变换及其证明方法,可较好地解决一些较复杂的、带约束条件的、涉及两个数列的对应项之积的和的上下界估计问题.

例1 已知$x_i\in 𝐑,i=1,2,\cdots ,n,n\ge 2$,满足$\sum _{i=1}^n\left|x_i\right|=1,\sum _{i=1}^n{x}_i=0$.

证明:$\left|\sum _{i=1}^n\frac{x_i}{i}\right|⩽\frac{1}{2}-\frac{1}{2n}$.
讲解:记诸$x_i$中全体非负数之和为A,全体负数之和为B,则由条件有A-B=1,且A+B=0.故必有A=$\frac{1}{2}$,B=$-\frac{1}{2}$.
记$S_k=\sum _{i=1}^k{x}_i,k=1,2,\cdots ,n,S_0=0,$则$\left|S_k\right|⩽\frac{1}{2},k=1,2,\cdots ,n$.
由阿贝尔变换有$$\sum _{i=1}^n\frac{x_i}{i}=\frac{1}{n}{S}_n+\sum _{i=1}^{n-1}{S}_i(\frac{1}{i}-\frac{1}{i+1})=\sum _{i=1}^{n-1}{S}_i(\frac{1}{i}-\frac{1}{i+1})$$从而,$$\left|\sum _{i=1}^n\frac{x_i}{i}\right|⩽\sum _{i=1}^{n-1}\left|S_i\right|(\frac{1}{i}-\frac{1}{i+1})⩽\frac{1}{2}\sum _{i=1}^{n-1}(\frac{1}{i}-\frac{1}{i+1})=\frac{1}{2}-\frac{1}{2n}$$

例2 设$x\in 𝐑,n\in 𝐍$.求证$\sum _{i=1}^n\frac{[ix]}{i}⩽[nx]$.这里$[x]$表示不超过$x$的最大整数.

讲解:从求证式的左边看,似可用$\sum _{i=1}^n\frac{[ix]}{i}=\frac{1}{n}\sum _{i=1}^n[ix]+\sum _{k=1}^{n-1}(\frac{1}{k}-\frac{1}{k+1})\sum _{i=1}^k[ix]$.但以下难以进行,关键是$\sum _{i=1}^n[ix]$与结论的关系不明显.转而用$$\sum _{i=1}^n[ix]=\sum _{i=1}^{n}i\cdot \frac{[ix]}{i}=n\sum _{i=1}^n\frac{[ix]}{i}+\sum _{k=1}^{n-1}(-1)\sum _{i=1}^k\frac{[ix]}{i}$$可推出$$n\sum _{i=1}^n\frac{[ix]}{i}=\sum _{i=1}^n[ix]+\sum _{k=1}^{n-1}\left(\sum _{i=1}^k\frac{[ix]}{i}\right)$$便为用数学归纳法证明此题扫清了障碍.最后的证明还要用到关系式$[x]+[y]\le [x+y]$.请读者自己完成.

例3 设$x_i\ge 0,i=1,2,\dots ,n$,且$\sum _{i=1}^n{x}_i^2+2\sum _{1⩽k<j⩽n}\sqrt{\frac{k}{j}}\cdot x_k{x}_j=1$.求$\sum _{i=1}^n{x}_i$的最大值和最小值.

讲解:易得$\sum _{i=1}^n{x}_i$的最小值为1(诸$x_i$中一个为1,而其余全为零时达到).为求最大值,注意到$$\begin{array}{rl}1& =\sum _{i=1}^{n}i{\left(\frac{x_i}{\sqrt{i}}\right)}^2+2\sum _{1⩽k<j⩽n}k(\frac{x_k}{\sqrt{k}}\cdot \frac{x_j}{\sqrt{j}})\\ & =\sum _{k=1}^{n}k\cdot \frac{x_k}{\sqrt{k}}(\frac{x_k}{\sqrt{k}}+2\sum _{i=k+1}^n\frac{x_i}{\sqrt{i}})\\ & =\sum _{k=1}^{n}k\cdot \frac{x_k}{\sqrt{k}}(\sum _{i=k}^n\frac{x_i}{\sqrt{i}}+\sum _{i=k+1}^n\frac{x_i}{\sqrt{i}})\\ & =\sum _{k=1}^{n}k(\sum _{i=k}^n\frac{x_i}{\sqrt{i}}-\sum _{i=k+1}^n\frac{x_i}{\sqrt{i}})(\sum _{i=k}^n\frac{x_i}{\sqrt{i}}+\sum _{i=k+1}^n\frac{x_i}{\sqrt{i}})\end{array}$$若令$y_i=\sum _{j=i}^n\frac{x_j}{\sqrt{j}},i=1,2,\cdots ,n$,则诸$y_i\ge 0$.逆用阿贝尔变换的证明方法可将条件化为$\sum _{i=1}^n{y}_i^2=1$.再由$y_i=\sum _{j=i}^n\frac{x_j}{\sqrt{j}},i=1,2,\cdots ,n$有$x_n=\sqrt{n}{y}_n,x_i=\sqrt{i}(y_i-y_{i+1}),i=1,2,\cdots ,n-1$.
记$y_{n+1}=0$,故有$\sum _{i=1}^n{x}_i=\sum _{i=1}^n\sqrt{i}(y_i-y_{i+1})=\sum _{i=1}^n(\sqrt{i}-\sqrt{i-1})y_i$.利用柯西不等式可得$\sum _{i=1}^n{x}_i$的最大值为$\sqrt{\sum _{i=1}^n(\sqrt{i}-\sqrt{i-1}{)}^2}$(当$x_i=\frac{2i-\sqrt{i^2+i}-\sqrt{i^2-i}}{\sqrt{\sum _{i=1}^n(\sqrt{i}-\sqrt{i-1}{)}^2}},i=1,2,\cdots ,n$时取到).

例4 实数$x_1,x_2,\cdots ,x_{2001}$满足$\sum _{k=1}^{2000}|x_k-x_{k+1}|=2001$,令$y_k=\frac{1}{k}(x_1+x_2+\cdots +x_k),k=1,2,\cdots ,2001$.求$\sum _{k=1}^{2000}|y_k-y_{k+1}|$的最大可能值.

讲解:由于不知$x_k$和$x_{k+1}$的大小关系,可将差$x_k-x_{k+1}$视为整体,将条件$\sum _{k=1}^{2000}|x_k-x_{k+1}|=2001$视为关于$x_k-x_{k+1}$的一个约束关系.作代换$a_0=x_1,a_k=x_{k+1}-x_k,k=1,2,\cdots ,2000$,则$x_1=a_0,x_k=\sum _{i=0}^{k-1}{a}_i,k=2,3,\cdots ,2001$,条件即为$\sum _{k=1}^{2000}\left|a_k\right|=2001$.此时$$\begin{array}{l}{y}_k=\frac{1}{k}[a_0+(a_0+a_1)+\cdots +\sum _{i=0}^{k-1}{a}_i]\\ =\frac{1}{k}[k{a}_0+(k-1)a_1+\cdots +a_{k-1}]\\ y_{k+1}=\frac{1}{k+1}[(k+1)a_0+k{a}_1+\cdots +2a_{k-1}+a_k]\end{array}$$则$$\begin{array}{l}|y_k-y_{k+1}|\\ =\frac{1}{k(k+1)}|-a_1-2a_2-\cdots -k{a}_k|\\ ⩽\frac{1}{k(k+1)}(\left|a_1\right|+2\left|a_2\right|+\cdots +k\left|a_k\right|)\end{array}$$记$A_k=\left|a_1\right|+2\left|a_2\right|+\cdots +k\left|a_k\right|,k=1,2,\cdots ,2000,A_0=0,$则$$\begin{array}{l}\sum _{k=1}^{2000}|y_k-y_{k+1}|⩽\sum _{k=1}^{2000}(\frac{1}{k}-\frac{1}{k+1})A_k\\ =\sum _{k=1}^{2000}\frac{1}{k}(A_k-A_{k-1})-\frac{1}{2001}\cdot A_{2000}\\ =\sum _{k=1}^{2000}\left|a_k\right|-\frac{1}{2001}\cdot A_{2000}\end{array}$$又$A_{2000}=\left|a_1\right|+2\left|a_2\right|+\cdots +2000·\left|a_{2000}\right|⩾\sum _{k=1}^{2000}\left|a_k\right|$,故$$\begin{array}{l}\sum _{k=1}^{2000}|y_k-y_{k+1}|\\ ⩽\sum _{k=1}^{2000}\left|a_k\right|-\frac{1}{2001}\cdot \sum _{k=1}^{2000}\left|a_k\right|\\ =\frac{2000}{2001}\cdot \sum _{k=1}^{2000}\left|a_k\right|=2000\end{array}$$由上述过程知,当且仅当$|a_1|=2001,a_2=a_3=\cdots =a_{2000}=0$时等号成立.故所求最大值为2000.

例5 已知$a_1,a_2,\cdots ,a_n$和$b_1,b_2,\cdots ,b_n$为实数.证明:使得对任何满足$x_1\le x_2\le \cdots \le x_n$的实数,不等式$\sum _{i=1}^n{a}_i{x}_i\le \sum _{i=1}^n{b}_i{x}_i$恒成立的充要条件是$\sum _{i=1}^k{a}_i⩾\sum _{i=1}^k{b}_i,k=1,2,\cdots ,n-1$,且$\sum _{i=1}^n{a}_i=\sum _{i=1}^n{b}_i$.

讲解:记$S_k=\sum _{i=1}^k{a}_i,T_k=\sum _{i=1}^k{b}_i,S_0=T_0=0$,则条件$\sum _{i=1}^n{a}_i{x}_i⩽\sum _{i=1}^n{b}_i{x}_i$可化为$$\begin{array}{lcr}\text{(1)}& S_n{x}_n+\sum _{k=1}^{n-1}{S}_k(x_k-x_{k+1})⩽T_n{x}_n+\sum _{k=1}^{n-1}{T}_k(x_k-x_{k+1})& \text{(1)}\end{array}$$取$x_1=x_2=\cdots =x_n$,有$S_n{x}_n⩽T_n{x}_n$,由$x_n$的任意性知必有$S_n=T_n$.
取$x_1=x_2=\cdots =x_k=-1(1⩽k⩽n-1)$,$x_{k+1}=\cdots =x_n=0$,可得$\sum _{i=1}^k{a}_i⩾\sum _{i=1}^k{b}_i$.必要性得证.
充分性只要求在“$S_n=T_n$,且$S_k\ge T_k(1\le k\le n-1)$”下证明式(1)成立.

例6 给定$c\in (\frac{1}{2},1)$求最小常数$M$使得对任意整数$n\ge 2$及实数$0<a_1\le a_2\le \dots \le a_n$,只要满足$$\begin{array}{lcr}\text{(1)}& \frac{1}{n}\sum _{k=1}^{n}k{a}_k=c\sum _{k=1}^n{a}_k& \text{(1)}\end{array}$$总有$\sum _{k=1}^n{a}_k⩽M\sum _{k=1}^m{a}_k$,其中$m=[cn]$表示不超过$cn$的最大整数.

讲解:应先据式(1)用特殊值法求出$M$的一个下界,最简单的方法是取诸$a_k$全相等.但由于$c$事先给定,诸$a_k$全相等时不一定能满足条件,因而先退一步,令$a_1=\dots =a_m$,而$a_m=\dots =a_n$,不妨设$a_1=1$,代入式(1)有$$\frac{m(m+1)}{2}+[\frac{n(n+1)}{2}-\frac{m(m+1)}{2}]a_{m+1}=cn[m+(n-m)a_{m+1}]$$由此可解出$a_{m+1}=\frac{m(2cn-m-1)}{(n-m)(n+m+1-2cn)}$(注意到$cn\ge m\ge 1$,且$cn<m+1\le n$).将取定的这组正数代入$\sum _{k=1}^n{a}_k⩽M\sum _{k=1}^m{a}_k$中,有$$m+(n-m)\cdot \frac{m(2cn-m-1)}{(n-m)(n+m+1-2cn)}\le mM$$则$M⩾1+\frac{2cn-m-1}{n+m+1-2cn}=\frac{n}{n+m+1-2cn}⩾\frac{n}{n+1-cn}=\frac{1}{1-c+\frac{1}{n}}.$令$n\to \infty$得$M⩾\frac{1}{1-c}$.欲证所求最小常数恰为$\frac{1}{1-c}$,应证对满足式(1)的任何递增数列$\{a_n\}$,恒有$$\begin{array}{lcr}\text{(2)}& \sum _{k=1}^n{a}_k⩽\frac{1}{1-c}\sum _{k=1}^m{a}_k& \text{(2)}\end{array}$$记$S_0=0,S_k=\sum _{i=1}^k{a}_i,k=1,2,\cdots ,n$,由阿贝尔变换得$$\begin{array}{lcr}\text{(3)}& (n-cn)S_n=S_1+S_2+\cdots +S_{n-1}& \text{(3)}\end{array}$$现要将$S_1,S_2,\dots ,S_n$的关系式(3)变为只含$S_m$和$S_n$的关系式(2),应设法用$S_m$和$S_n$来表示诸$S_k$,或限制其范围.显然$k\le m$时,有$S_k\le S_m$,但若将$S_1,S_2,\cdots ,S_{m-1}$都直接放大到$S_m$就可能过头了,根本不需要$\{a_n\}$的递增条件.而由$\{a_n\}$的递增条件,当$k\le m$时,前$k$个数的平均数不超过前$m$个数的平均数,即$S_k⩽\frac{k}{m}\cdot S_m$.
又$m+1⩽k⩽n$时,$S_k=S_m+a_{m+1}+\cdots +a_k$.同样由$\{a_n\}$的递增条件,$k-m$个数$a_{m+1},\cdots ,a_k$的平均数不超过$n-m$个数$a_{m+1},\cdots ,a_n$的平均数.于是,$m+1⩽k⩽n$时,有$$S_k⩽S_m+\frac{k-m}{n-m}(a_{m+1}+\cdots +a_n)=\frac{n-k}{n-m}\cdot S_m+\frac{k-m}{n-m}\cdot S_n$$式(3)化为$$\begin{gathered} (n-cn)S_n \\ ⩽\frac{1+\cdots +m}{m}\cdot S_m+\frac{(n-m-1)+\cdots +1}{n-m}\cdot S_m+ \\ \frac{1+\cdots +(n-1-m)}{n-m}\cdot S_n \\ =\frac{n}{2}\cdot S_m+\frac{n-m-1}{2}\cdot S_n \end{gathered}$$故$$S_n⩽\frac{n}{n+1+m-2cn}\cdot S_m<\frac{n}{n-cn}\cdot S_m=\frac{1}{1-c}\cdot S_m$$

练习题

1. (阿贝尔不等式)设$a_k,b_k\in 𝐑(k=1,2,\dots ,n),b_1⩾b_2⩾\cdots ⩾b_n⩾0$,对$k=1,2,\cdots ,n,$记$S_k=\sum _{i=1}^k{a}_i,M=\underset{1\le k<n}{max}{S}_k,m=\underset{1<k<n}{min}{S}_k.$则有$m{b}_1⩽\sum _{k=1}^n{a}_k{b}_k⩽M{b}_1$
2. 已知$a_1,a_2,\cdots ,a_n$为任意两两各不相同的正整数.求证:对任意正整数$n$,下列不等式成立:$$\sum _{k=1}^n\frac{a_k}{k^2}⩾\sum _{k=1}^n\frac{1}{k}$$(提示:由阿贝尔变换得$\sum _{k=1}^n\frac{a_k}{k^2}=\frac{1}{n^2}{S}_n+\sum _{k=1}^{n-1}[\frac{1}{k^2}-\frac{1}{(k+1{)}^2}]S_k$,其中$S_k=\sum _{i=1}^k{a}_i⩾\sum _{i=1}^{k}i$.)
3. (钟开莱不等式)设$a_i,b_i\in 𝐑(k=1,2,\dots ,n),a_1⩾a_2⩾\cdots ⩾a_n⩾0$,对$k=1,2,\cdots ,n$恒有$\sum _{i=1}^k{a}_i⩽\sum _{i=1}^k{b}_i$.则必有$\sum _{i=1}^n{a}_i^2⩽\sum _{i=1}^n{b}_i^2$.
   (提示:先用阿贝尔变换证明$\sum _{i=1}^n{a}_i^2⩽\sum _{i=1}^n{a}_i{b}_i$,再用柯西不等式推出结论.)
4. 已知$a_1,a_2,\cdots ,a_n,\cdots$是实数列,满足$a_{i+j}\le a_i+a_j(i,j\in {𝐍}^{\ast })$[这称为Subadditivity],证明:
   (1)$a_n⩽\frac{2}{n-1}\sum _{i=1}^{n-1}{a}_i(n⩾2,n\in 𝐍)$;
   (2)$a_1+\frac{a_2}{2}+\frac{a_3}{3}+\cdots +\frac{a_n}{n}⩾a_n$
   (提示:(1)$2\sum _{i=1}^{n-1}{a}_i=\sum _{i=1}^{n-1}(a_i+a_{n-i})\ge (n-1)a_n$;(2)仿(1)得$a_k⩽\frac{2}{k-1}\sum _{i=1}^{k-1}{a}_i$再对求证式左边用阿贝尔变换.)
5. 设$a_1⩾a_2⩾\cdots ⩾a_n⩾0,b_1⩾a_1,b_1{b}_2⩾a_1{a}_2,\cdots ,b_1{b}_2\cdots b_n⩾a_1{a}_2\cdots a_n$.求证$b_1+b_2+\cdots +b_n⩾a_1+a_2+\cdots +a_n$.
   (提示:令$c_i=\frac{b_i}{a_i}(1⩽i⩽n)$,结论转化为$\sum _{i=1}^n(c_i-1)a_i⩾0$,用阿贝尔变换及均值不等式可得).

应用阿贝尔变换解竞赛题. 方廷刚.《中等数学》2003