阿贝尔分布求和法的应用(一)

1. (和差变换公式)设$m<n$.则
$$\sum_{k=m}^{n}(A_{k}-A_{k-1})b_{k}=A_{n}b_{n}-A_{m-1}b_{m}+\sum_{k=m}^{n-1}A_{k}(b_{k}-b_{k+1})$$
证明：直接计算即可。
\begin{align*}
\sum_{k=m}^{n}(A_{k}-A_{k-1})b_{k}&=\sum_{k=m}^{n}A_{k}b_{k}-\sum_{k=m}^{n}A_{k-1}b_{k}\\
&=\sum_{k=m}^{n}A_{k}b_{k}-\sum_{m-1}^{n-1}A_{k}b_{k+1}\\
&=(A_{n}b_{n}-A_{m-1}b_{m})+\sum_{k=m}^{n-1}A_{k}(b_{k}-b_{k+1})
\end{align*}
2. (分部求和法)设$s_{k}=a_{1}+a_{2}+\cdots+a_{k},(k=1,2,\cdots,n)$.则
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})$$
证明：补充定义$s_{0}=0$,利用第一题的结论即可。令本命题和第一题等价。
不妨设$m<n$,由题知
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})$$
$$\sum_{k=1}^{m-1}a_{k}b_{k}=s_{m-1}b_{m-1}+\sum_{k=1}^{m-2}s_{k}(b_{k}-b_{k+1})$$
两式相减得
$$\sum_{k=m}^{n}a_{k}b_{k}=s_{n}b_{n}-s_{m-1}b_{m-1}+\sum_{k=m-1}^{n-1}s_{k}(b_{k}-b_{k+1})$$\\
3. 设$s_{n}=a_{1}+a_{2}+\cdots+a_{n}\to s(n\to \infty)$,则
$$\sum_{k=1}^{n}a_{k}b_{k}=sb_{1}+(s_{n}-s)b_{n}-\sum_{k=1}^{n-1}(s_{k}-s)(b_{k+1}-b_{k})$$
证明：由分布求和知
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}-\sum_{k=1}^{n-1}s_{k}(b_{k+1}-b_{k})$$
而
$$s(b_{n}-b_{1})=s\sum_{k=1}^{n-1}(b_{k+1}-b_{k})$$
两式相减即得结论。
4. (阿贝耳引理)若对一切$n=1,2,3,\cdots$而言
$$b_{1}\geq b_{2}\geq \cdots \geq b_{n}\geq 0$$
$$m\leq a_{1}+a_{2}+\cdot+a_{n}\leq M$$
则有
$$b_{1}m\geq a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}\leq b_{1}M$$
证明：设$s_{k}=a_{1}+a_{2}+\cdots+a_{k},(k=1,2,\cdots,n)$.由于$b_{k}\geq 0,b_{k}-b_{k+1}\geq 0$则
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})\leq b_{n}M+M\sum_{k=1}^{n-1}(b_{k}-b_{k+1})=b_{1}M$$
左边不等式证明类似
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})\geq b_{n}m+m\sum_{k=1}^{n-1}(b_{k}-b_{k+1})=b_{1}m$$
5. 设$a_{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$为任意实数或复数，又设
$$A=\max \{|a_{1}|,|a_{1}+a_{2}|,\cdots,|a_{1}+a_{2}+\cdots+a_{n}|\}$$
则
$$\left|\sum_{k=1}^{n}a_{k}b_{k}\right|\leq A \left\{\sum_{k=1}^{n-1}|b_{k+1}-b_{k}|+|b_{n}|\right\}$$
证明：使用绝对值不等式放缩即可。
$$\left|\sum_{k=1}^{n}a_{k}b_{k}\right|\leq |s_{n}|\times|b_{n}|+\sum_{k=1}^{n-1}|s_{k}|\times |b_{k}-b_{k+1}|\leq A \left\{\sum_{k=1}^{n-1}|b_{k+1}-b_{k}|+|b_{n}|\right\}$$
6. (Kronecker) 设$\varphi(n)>0,\varphi(n)\uparrow \infty(n \to \infty)$.又设$\sum_{n=1}^{\infty}a_{n}$收敛. 则
$$\sum_{k=1}^{n}a_{k}\varphi(k)=o(\varphi(n)),(n\to \infty)$$
证明：设$s_{n}\to s (n\to \infty)$那么对任意$\varepsilon>0$存在$m>0,n>m$时
$$|s_{n}-s|<\varepsilon$$
由分部求和法知(重复第二题第三题)步骤
\begin{align*}
\sum_{k=1}^{n}a_{k}b_{k}&=s\varphi(1)+(s_{n}-s)\varphi(n)-\sum_{k=1}^{n-1}(s_{k}-s)(\varphi(k+1)-\varphi(k))\\
&=O(1)+o(\varphi(n))-\sum_{k=1}^{m}(s_{k}-s)(\varphi(k+1)-\varphi(k))-\sum_{k=m+1}^{n-1}(s_{k}-s)(\varphi(k+1)-\varphi(k))\\
&\leq O(1)+o(\varphi(n))+\varepsilon (\varphi(n)-\varphi(m+1))
\end{align*}
由于$\varphi(n)\to \infty$,知$\sum_{k=1}^{n}a_{k}\varphi(k)=o(\varphi(n)),(n\to \infty)$
7. 设$\varphi(n)\downarrow 0(n \to \infty)$,且$\sum_{n=1}^{\infty}a_{n}\varphi(n)$为收敛，则
$$\lim_{n\to\infty}(a_{1}+a_{2}+\cdots+a_{n})\varphi(n)=0$$
证明：将$a_{n}\varphi(n)$看作命题6中的$a_{n}$,将$\varphi^{-1}(n)$看作命题6中的$\varphi(n)$,命题得证。
\textbf{另证}: 直接利用阿贝耳引理证明.任意给定$\varepsilon>0$,由Cauchy收敛准则知,存在$N$,$n>N$时
$$-\frac{\varepsilon}{2}<a_{N}\varphi(N)+a_{N+1}\varphi{N+1}+\cdots+a_{n}\varphi(n)<\frac{\varepsilon}{2}$$
又有单调递减非负列
$$\varphi^{-1}(n)\geq \varphi^{-1}(n-1)\geq \cdots\geq\varphi^{-1}(N)>0$$
重新定义序列,使用阿贝耳引理命题4,即得
$$|a_{N}+a_{N+1}+\cdot+a_{n}|<\varphi^{-1}(n)\frac{\varepsilon}{2}$$
即
$$|(a_{N}+a_{N+1}+\cdot+a_{n})\varphi(n)|<\varphi^{-1}(n)\frac{\varepsilon}{2}$$
由于$\varphi(n)\to 0,n\to \infty$
$$\limsup|(a_{1}+\cdots+a_{N-1}+a_{N}+\cdots+a_{n})\varphi(n)|\leq 0+\frac{\varepsilon}{2}$$
命题得证.
8.(Dirichlet) 设$\sigma>0$,则下列的狄利克雷级数
$$a_{1}\cdot 1^{-\sigma}+a_{2}\cdot 2^{-\sigma}+a_{2}\cdot 3^{-\sigma}+\cdots+a_{n}\cdot n^{-\sigma}+\cdots$$
收敛时，必有
$$\lim_{n\to\infty}(a_{1}+a_{2}+\cdots+a_{n})n^{-\sigma}=0$$
证明：此命题是命题7的特例。
9. 设$\{z_{n}\}_{1}^{\infty}$为任意一个复数列而
$$\sum_{n=1}^{\infty}|z_{n+1}^{-1}-z_{n}^{-1}|=\infty$$
又设级数$\sum_{n=1}^{\infty}a_{n}z_{n}$为收敛，则必有
$$\lim_{N\to\infty}\left(\sum_{n=1}^{N}a_{n}\right)\left(\sum_{n=1}^{N}|z_{n+1}^{-1}-z_{n}^{-1}|\right)^{-1}=0$$
证明:此类题目都是对和式$|\sum a_{n}|$作估计,设$s_{n}=\sum_{k=1}^{n}a_{k}z_{k},s_{n}\to s (n\to \infty)$, 则有估计
\begin{align*}
\left|\sum_{n=1}^{N}a_{n}\right|&=\left|\sum_{n=1}^{N}\left(a_{n}z_{n}\right)z_{n}^{-1}\right|\\
&=\left|\sum_{n=1}^{N}(s_{n}-s_{n-1})z_{n}^{-1}\right|\\
&=\left|\sum_{n=1}^{N}s_{n}z_{n}^{-1}-\sum_{n=1}^{N}s_{n-1}z_{n}^{-1}\right|\\
&=\left|s_{N}z_{N}^{-1}+\sum_{n=1}^{N-1}s_{n}(z_{n}^{-1}-z_{n+1}^{-1})\right|\\
&=\left|s_{N}z_{N+1}^{-1}+\sum_{n=1}^{N}s_{n}(z_{n}^{-1}-z_{n+1}^{-1})\right|\\
&=\left|(s_{N}-s)z_{N+1}^{-1}+s z_{1}^{-1}+\sum_{n=1}^{N}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})\right|
\end{align*}
由$\lim_{N\to \infty}s_{N}-s=0$知,存在$m$,当$N>m$时,$|s_{n}-s|<\varepsilon$
因此,上述等式右边可放缩
\begin{align*}
R&\leq \left|(s_{N}-s)z_{N+1}^{-1})\right|+\left|sz_{1}^{-1}\right|+\left|\sum_{n=1}^{m}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})\right|+\left|\sum_{n=m}^{N}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})\right|\\
&\leq\varepsilon |z_{N+1}^{-1}|+\left|sz_{1}^{-1}\right|+\varepsilon \left|z_{n}^{-1}-z_{n+1}^{-1}\right|+\left|\sum_{n=1}^{m}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})\right|\\
&=\varepsilon |z_{N+1}^{-1}|+\varepsilon\sum_{m}^{N}\left|z_{n}^{-1}-z_{n+1}^{-1}\right|+M
\end{align*}
因此,
\begin{align*}
\frac{\left|\sum_{n=1}^{N}a_{n}\right|}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}&\leq\varepsilon \cdot \frac{|z_{n+1}^{-1}|}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}+\frac{M}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}+\varepsilon \cdot \frac{\sum_{m}^{N}\left|z_{n}^{-1}-z_{n+1}^{-1}\right|}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}\\
&\leq \varepsilon+\frac{M}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}+\varepsilon
\end{align*}
从而
$$\limsup \frac{\left|\sum_{n=1}^{N}a_{n}\right|}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}\leq 2\varepsilon$$
由于$\varepsilon$为任意数
$$\limsup \frac{\left|\sum_{n=1}^{N}a_{n}\right|}{\sum_{n=1}^{N}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|}=0$$
证明方法: 阿贝尔分部求和法+分段估计+上极限。
10. 设当$k=1,2,3,\cdots$时，有
$$b_{k}\geq b_{k+1},\frac{1}{2}(b_{k}+b_{k+1})\geq b_{k+1}$$
并且
$$m\leq s_{1}+s_{2}+\cdots+s_{k}\leq M$$
其中$s_{k}=a_{1}+a_{2}+\cdots+a_{k}$.则有下列不等式成立
$$m(b_{1}-b_{2})+s_{n}b_{n}<\sum_{k=1}^{n}a_{k}b_{k}<M(b_{1}-b_{2})+s_{n}b_{n}$$
证明: 设
$$M_{n}=\sum_{k=1}^{n}s_{k},\, \tilde{b}_{k}=b_{k}-b_{k+1}$$
由$b_{k}+b_{k+2}\geq 2b_{k+1}$易知$\tilde{b}_{k}\geq \tilde{b}_{k+1}$.由分布求和公式
\begin{align*}
\sum_{k=1}^{n}a_{k}b_{k}&=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})\\
&=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}\tilde{b}_{k}\\
&=s_{n}b_{n}+M_{n-1}\tilde{b}_{n-1}+\sum_{k=1}^{n-2}M_{k}(\tilde{b}_{k}-\tilde{b}_{k+1})\\
&\leq s_{n}b_{n}+M\tilde{b}_{n-1}+M\sum_{k=1}^{n-2}(\tilde{b}_{k}-\tilde{b}_{k+1})\\
&=s_{n}b_{n}+(b_{1}-b_{2})M
\end{align*}
左边不等式证明类似。
11.设$N$为一固定的大整数,$a_{1},a_{2},\cdots,a_{N},b_{1},b_{2},\cdots,b_{N}$为任意两组常数. 今定义$b_{k}=0(k>N)$以及
$$\Delta^{m}b_{k}=\Delta^{m-1}b_{k+1}-\Delta^{m-1}b_{k},\, \Delta b_{k}=b_{k+1}-b_{k}$$
$$s_{k}^{(m)}=\sum_{\nu=1}^{k}s_{\nu}^{(m-1)},s_{k}^{(1)}=a_{1}+a_{2}+\cdots+a_{k}$$
则有下列恒等式成立:
$$\sum_{k=1}^{N}a_{k}b_{k}=(-1)^{m}\sum_{k=1}^{N}s_{k}^{(m)}\Delta^{m}b_{k}$$
证明：反复利用分布求和公式可得
\begin{align*}
\sum_{k=1}^{N}a_{k}b_{k}&=s_{N}b_{N}+\sum_{k=1}^{N-1}s_{k}\left(-\Delta b_{k}\right)\\
&=s_{N}b_{N}+s_{N-1}^{(2)}\left(-\Delta b_{N-1}\right)+\sum_{k=1}^{N-2}s_{k}^{2}\left(\Delta^{2} b_{k}\right)\\
&=s_{N}^{(1)}b_{N}+s_{N-1}^{(2)}\left(-\Delta b_{N-1}\right)+s_{N-2}^{(3)}\left(\Delta^{2} b_{N-2}\right)+\cdots+(-1)^{N-1}s_{1}^{(N)}\Delta^{N-1}b_{1}\\
&=\sum_{k=1}^{N}(-1)^{k-1}s_{N-k+1}^{(k)}\Delta^{k-1}b_{N-k+1}
\end{align*}

11.设$N$为一固定的大整数,$a_{1},a_{2},\cdots,a_{N},b_{1},b_{2},\cdots,b_{N}$为任意两组常数. 今定义$b_{k}=0(k>N)$以及
$$\Delta^{m}b_{k}=\Delta^{m-1}b_{k+1}-\Delta^{m-1}b_{k},\, \Delta b_{k}=b_{k+1}-b_{k}$$
$$s_{k}^{(m)}=\sum_{\nu=1}^{k}s_{\nu}^{(m-1)},s_{k}^{(1)}=a_{1}+a_{2}+\cdots+a_{k}$$
则有下列恒等式成立:
$$\sum_{k=1}^{N}a_{k}b_{k}=(-1)^{m}\sum_{k=1}^{N}s_{k}^{(m)}\Delta^{m}b_{k}$$
证明：反复利用分布求和公式可得
\begin{align*}
\sum\limits_{k=1}^{N}a_{k}b_{k}&=s_{N}b_{N}+\sum\limits_{k=1}^{N-1}s_{k}\left(-\Delta b_{k}\right)\\
&=s_{N}b_{N}+s_{N-1}^{(2)}\left(-\Delta b_{N-1}\right)+\sum\limits_{k=1}^{N-2}s_{k}^{2}\left(\Delta^{2} b_{k}\right)\\
&=s_{N}^{(1)}b_{N}+s_{N-1}^{(2)}\left(-\Delta b_{N-1}\right)+s_{N-2}^{(3)}\left(\Delta^{2} b_{N-2}\right)+\cdots+(-1)^{N-1}s_{1}^{(N)}\Delta^{N-1}b_{1}\\
&=\sum\limits_{k=1}^{N}(-1)^{k-1}s_{N-k+1}^{(k)}\Delta^{k-1}b_{N-k+1}
\end{align*}
12. 设$a_{k}>0,b_{k}>0,$而$\{v_{k}\}$为单调下降的正数列. 又设
$$H=\max\left(\frac{B_{0}}{A_{0}},\frac{B_{1}}{A_{1}},\cdots,\frac{B_{n}}{A_{n}}\right),\,\,h=\min\left(\frac{B_{0}}{A_{0}},\frac{B_{1}}{A_{1}},\cdots,\frac{B_{n}}{A_{n}}\right) $$
$$H_{m}=\max\left(\frac{B_{m}}{A_{m}},\frac{B_{m+1}}{A_{m+1}},\cdots,\frac{B_{n}}{A_{n}}\right),\,\,h_{m}=\min\left(\frac{B_{m}}{A_{m}},\frac{B_{m+1}}{A_{m+1}},\cdots,\frac{B_{n}}{A_{n}}\right)$$
此处$A_{k}=a_{1}+a_{2}+\cdots+a_{k},\, B_{k}=b_{1}+b_{2}+\cdots+b_{k}.$ 则有下列不等式成立:
$$h_{m}+(h-h_{m})\frac{\sum\limits_{k=0}^{m}a_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}\leq \frac{\sum\limits_{k=0}^{n}b_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}\leq H_{m}+(H-H_{m})\frac{\sum\limits_{k=0}^{m}a_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}$$
证明: 分部求和法+分段估计, 只证右边不等式左边不等式证明类似.
\begin{align*}
\frac{\sum\limits_{k=0}^{n}b_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}-H_{m}&=\frac{\sum\limits_{k=0}^{m-1}B_{k}(v_{k}-v_{k+1})+\sum\limits_{k=m}^{n}B_{k}(v_{k}-v_{k+1})+B_{n}v_{n}}{\sum\limits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}-H_{m}\\
&\leq \frac{H\sum\limits_{k=0}^{m-1}A_{k}(v_{k}-v_{k+1})+H_{m}\sum\limits_{k=m}^{n}A_{k}(v_{k}-v_{k+1})+H_{m}A_{n}v_{n}}{\sum\limits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}-H_{m}\\
&=\frac{(H-H_{m})\sum\limits_{k=0}^{m-1}A_{k}(v_{k}-v_{k+1})}{\sum\limits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}\\
&\leq (H-H_{m})\frac{\sum\limits_{k=0}^{m}a_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}
\end{align*}
13. 保留命题12的全部假设，但将$\{v_{n}\}$改为单调上升的的数列. 则有
$$H_{m}-\frac{(H_{m}-h_{m})A_{n}v_{n}+(H-H_{m})A_{m}v_{m}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}\leq \frac{\sum\limits_{k=0}^{n}b_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}\leq h_{m}+\frac{(H_{m}-h_{m})A_{n}v_{n}+(h_{m}-h)A_{m}v_{m}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}$$
证明: 分部求和法+分段估计, 只证右边不等式左边不等式证明类似.
\begin{align*}
\frac{\sum\limits_{k=0}^{n}b_{k}v_{k}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}-h_{m}&=\frac{\sum\limits_{k=0}^{m-1}B_{k}(v_{k}-v_{k+1})+\sum\limits_{k=m}^{n}B_{k}(v_{k}-v_{k+1})+B_{n}v_{n}}{\sum\limits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}-h_{m}\\
&=\frac{\sum\limits_{k=0}^{m-1}(B_{k}-h_{m}A_{k})(v_{k}-v_{k+1})+\sum\limits_{k=m}^{n}(B_{k}-h_{m}A_{k})(v_{k}-v_{k+1})+(B_{n}-h_{m}A_{n})v_{n}}{\sum\limits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}\\
&\leq\frac{(H_{m}-h_{m})A_{n}v_{n}+(h_{m}-h)A_{m}v_{m}}{\sum\limits_{k=0}^{n}a_{k}v_{k}}
\end{align*}