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。
egin{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$s_{n}=a_{1}+a_{2}+cdots+a_{n} o s(n o infty)$,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
$$s(b_{n}-b_{1})=ssum_{k=1}^{n-1}(b_{k+1}-b_{k})$$
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
$$b_{1}geq b_{2}geq cdots geq b_{n}geq 0$$
$$mleq a_{1}+a_{2}+cdot+a_{n}leq M$$
则有
$$b_{1}mgeq a_{1}b_{1}+a_{2}b_{2}+cdots+a_{n}b_{n}leq b_{1}M$$
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
$$sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})leq b_{n}M+Msum_{k=1}^{n-1}(b_{k}-b_{k+1})=b_{1}M$$
左边不等式证明类似
$$sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})geq b_{n}m+msum_{k=1}^{n-1}(b_{k}-b_{k+1})=b_{1}m$$
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
$$left|sum_{k=1}^{n}a_{k}b_{k}
ight|leq A left{sum_{k=1}^{n-1}|b_{k+1}-b_{k}|+|b_{n}|
ight}$$
证明：使用绝对值不等式放缩即可。
$$left|sum_{k=1}^{n}a_{k}b_{k}
ight|leq |s_{n}| imes|b_{n}|+sum_{k=1}^{n-1}|s_{k}| imes |b_{k}-b_{k+1}|leq A left{sum_{k=1}^{n-1}|b_{k+1}-b_{k}|+|b_{n}|
ight}$$
6. (Kronecker) 设$varphi(n)>0,varphi(n)uparrow infty(n o infty)$.又设$sum_{n=1}^{infty}a_{n}$收敛. 则
$$sum_{k=1}^{n}a_{k}varphi(k)=o(varphi(n)),(n o infty)$$
证明：设$s_{n} o s (n o infty)$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
egin{align*}
sum_{k=1}^{n}a_{k}b_{k}&=svarphi(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$varphi(n) o infty$,知$sum_{k=1}^{n}a_{k}varphi(k)=o(varphi(n)),(n o infty)$
7. 设$varphi(n)downarrow 0(n o infty)$,且$sum_{n=1}^{infty}a_{n}varphi(n)$为收敛，则
$$lim_{n oinfty}(a_{1}+a_{2}+cdots+a_{n})varphi(n)=0$$
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。
extbf{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
$$varphi^{-1}(n)geq varphi^{-1}(n-1)geq cdotsgeqvarphi^{-1}(N)>0$$
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$varphi(n) o 0,n o infty$
$$limsup|(a_{1}+cdots+a_{N-1}+a_{N}+cdots+a_{n})varphi(n)|leq 0+frac{varepsilon}{2}$$
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
$$lim_{n oinfty}(a_{1}+a_{2}+cdots+a_{n})n^{-sigma}=0$$
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
$$lim_{N oinfty}left(sum_{n=1}^{N}a_{n}
ight)left(sum_{n=1}^{N}|z_{n+1}^{-1}-z_{n}^{-1}|
ight)^{-1}=0$$
证明:此类题目都是对和式$|sum a_{n}|$作估计,设$s_{n}=sum_{k=1}^{n}a_{k}z_{k},s_{n} o s (n o infty)$, 则有估计
egin{align*}
left|sum_{n=1}^{N}a_{n}
ight|&=left|sum_{n=1}^{N}left(a_{n}z_{n}
ight)z_{n}^{-1}
ight|\
&=left|sum_{n=1}^{N}(s_{n}-s_{n-1})z_{n}^{-1}
ight|\
&=left|sum_{n=1}^{N}s_{n}z_{n}^{-1}-sum_{n=1}^{N}s_{n-1}z_{n}^{-1}
ight|\
&=left|s_{N}z_{N}^{-1}+sum_{n=1}^{N-1}s_{n}(z_{n}^{-1}-z_{n+1}^{-1})
ight|\
&=left|s_{N}z_{N+1}^{-1}+sum_{n=1}^{N}s_{n}(z_{n}^{-1}-z_{n+1}^{-1})
ight|\
&=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})
ight|
end{align*}
由$lim_{N o infty}s_{N}-s=0$知,存在$m$,当$N>m$时,$|s_{n}-s|<varepsilon$
因此,上述等式右边可放缩
egin{align*}
R&leq left|(s_{N}-s)z_{N+1}^{-1})
ight|+left|sz_{1}^{-1}
ight|+left|sum_{n=1}^{m}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})
ight|+left|sum_{n=m}^{N}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})
ight|\
&leqvarepsilon |z_{N+1}^{-1}|+left|sz_{1}^{-1}
ight|+varepsilon left|z_{n}^{-1}-z_{n+1}^{-1}
ight|+left|sum_{n=1}^{m}(s_{n}-s)(z_{n}^{-1}-z_{n+1}^{-1})
ight|\
&=varepsilon |z_{N+1}^{-1}|+varepsilonsum_{m}^{N}left|z_{n}^{-1}-z_{n+1}^{-1}
ight|+M
end{align*}
因此,
egin{align*}
frac{left|sum_{n=1}^{N}a_{n}
ight|}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}&leqvarepsilon cdot frac{|z_{n+1}^{-1}|}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}+frac{M}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}+varepsilon cdot frac{sum_{m}^{N}left|z_{n}^{-1}-z_{n+1}^{-1}
ight|}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}\
&leq varepsilon+frac{M}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}+varepsilon
end{align*}
从而
$$limsup frac{left|sum_{n=1}^{N}a_{n}
ight|}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}leq 2varepsilon$$
由于$varepsilon$为任意数
$$limsup frac{left|sum_{n=1}^{N}a_{n}
ight|}{sum_{n=1}^{N}left|z_{n+1}^{-1}-z_{n}^{-1}
ight|}=0$$
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
$$mleq s_{1}+s_{2}+cdots+s_{k}leq M$$
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
$$M_{n}=sum_{k=1}^{n}s_{k},\, ilde{b}_{k}=b_{k}-b_{k+1}$$
由$b_{k}+b_{k+2}geq 2b_{k+1}$易知$ ilde{b}_{k}geq ilde{b}_{k+1}$.由分布求和公式
egin{align*}
sum_{k=1}^{n}a_{k}b_{k}&=s_{n}b_{n}+sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})\
&=s_{n}b_{n}+sum_{k=1}^{n-1}s_{k} ilde{b}_{k}\
&=s_{n}b_{n}+M_{n-1} ilde{b}_{n-1}+sum_{k=1}^{n-2}M_{k}( ilde{b}_{k}- ilde{b}_{k+1})\
&leq s_{n}b_{n}+M ilde{b}_{n-1}+Msum_{k=1}^{n-2}( ilde{b}_{k}- ilde{b}_{k+1})\
&=s_{n}b_{n}+(b_{1}-b_{2})M
end{align*}
左边不等式证明类似。
11.设$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
$$Delta^{m}b_{k}=Delta^{m-1}b_{k+1}-Delta^{m-1}b_{k},\, Delta b_{k}=b_{k+1}-b_{k}$$
$$s_{k}^{(m)}=sum_{
u=1}^{k}s_{
u}^{(m-1)},s_{k}^{(1)}=a_{1}+a_{2}+cdots+a_{k}$$
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
egin{align*}
sum_{k=1}^{N}a_{k}b_{k}&=s_{N}b_{N}+sum_{k=1}^{N-1}s_{k}left(-Delta b_{k}
ight)\
&=s_{N}b_{N}+s_{N-1}^{(2)}left(-Delta b_{N-1}
ight)+sum_{k=1}^{N-2}s_{k}^{2}left(Delta^{2} b_{k}
ight)\
&=s_{N}^{(1)}b_{N}+s_{N-1}^{(2)}left(-Delta b_{N-1}
ight)+s_{N-2}^{(3)}left(Delta^{2} b_{N-2}
ight)+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
$$Delta^{m}b_{k}=Delta^{m-1}b_{k+1}-Delta^{m-1}b_{k},\, Delta b_{k}=b_{k+1}-b_{k}$$
$$s_{k}^{(m)}=sum_{
u=1}^{k}s_{
u}^{(m-1)},s_{k}^{(1)}=a_{1}+a_{2}+cdots+a_{k}$$
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
egin{align*}
sumlimits_{k=1}^{N}a_{k}b_{k}&=s_{N}b_{N}+sumlimits_{k=1}^{N-1}s_{k}left(-Delta b_{k}
ight)\
&=s_{N}b_{N}+s_{N-1}^{(2)}left(-Delta b_{N-1}
ight)+sumlimits_{k=1}^{N-2}s_{k}^{2}left(Delta^{2} b_{k}
ight)\
&=s_{N}^{(1)}b_{N}+s_{N-1}^{(2)}left(-Delta b_{N-1}
ight)+s_{N-2}^{(3)}left(Delta^{2} b_{N-2}
ight)+cdots+(-1)^{N-1}s_{1}^{(N)}Delta^{N-1}b_{1}\
&=sumlimits_{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
$$H=maxleft(frac{B_{0}}{A_{0}},frac{B_{1}}{A_{1}},cdots,frac{B_{n}}{A_{n}}
ight),\,\,h=minleft(frac{B_{0}}{A_{0}},frac{B_{1}}{A_{1}},cdots,frac{B_{n}}{A_{n}}
ight) $$
$$H_{m}=maxleft(frac{B_{m}}{A_{m}},frac{B_{m+1}}{A_{m+1}},cdots,frac{B_{n}}{A_{n}}
ight),\,\,h_{m}=minleft(frac{B_{m}}{A_{m}},frac{B_{m+1}}{A_{m+1}},cdots,frac{B_{n}}{A_{n}}
ight)$$
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:
$$h_{m}+(h-h_{m})frac{sumlimits_{k=0}^{m}a_{k}v_{k}}{sumlimits_{k=0}^{n}a_{k}v_{k}}leq frac{sumlimits_{k=0}^{n}b_{k}v_{k}}{sumlimits_{k=0}^{n}a_{k}v_{k}}leq H_{m}+(H-H_{m})frac{sumlimits_{k=0}^{m}a_{k}v_{k}}{sumlimits_{k=0}^{n}a_{k}v_{k}}$$
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.
egin{align*}
frac{sumlimits_{k=0}^{n}b_{k}v_{k}}{sumlimits_{k=0}^{n}a_{k}v_{k}}-H_{m}&=frac{sumlimits_{k=0}^{m-1}B_{k}(v_{k}-v_{k+1})+sumlimits_{k=m}^{n}B_{k}(v_{k}-v_{k+1})+B_{n}v_{n}}{sumlimits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}-H_{m}\
&leq frac{Hsumlimits_{k=0}^{m-1}A_{k}(v_{k}-v_{k+1})+H_{m}sumlimits_{k=m}^{n}A_{k}(v_{k}-v_{k+1})+H_{m}A_{n}v_{n}}{sumlimits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}-H_{m}\
&=frac{(H-H_{m})sumlimits_{k=0}^{m-1}A_{k}(v_{k}-v_{k+1})}{sumlimits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}\
&leq (H-H_{m})frac{sumlimits_{k=0}^{m}a_{k}v_{k}}{sumlimits_{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
$$H_{m}-frac{(H_{m}-h_{m})A_{n}v_{n}+(H-H_{m})A_{m}v_{m}}{sumlimits_{k=0}^{n}a_{k}v_{k}}leq frac{sumlimits_{k=0}^{n}b_{k}v_{k}}{sumlimits_{k=0}^{n}a_{k}v_{k}}leq h_{m}+frac{(H_{m}-h_{m})A_{n}v_{n}+(h_{m}-h)A_{m}v_{m}}{sumlimits_{k=0}^{n}a_{k}v_{k}}$$
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.
egin{align*}
frac{sumlimits_{k=0}^{n}b_{k}v_{k}}{sumlimits_{k=0}^{n}a_{k}v_{k}}-h_{m}&=frac{sumlimits_{k=0}^{m-1}B_{k}(v_{k}-v_{k+1})+sumlimits_{k=m}^{n}B_{k}(v_{k}-v_{k+1})+B_{n}v_{n}}{sumlimits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}-h_{m}\
&=frac{sumlimits_{k=0}^{m-1}(B_{k}-h_{m}A_{k})(v_{k}-v_{k+1})+sumlimits_{k=m}^{n}(B_{k}-h_{m}A_{k})(v_{k}-v_{k+1})+(B_{n}-h_{m}A_{n})v_{n}}{sumlimits_{k=0}^{n}A_{k}(v_{k}-v_{k+1})+A_{n}v_{n}}\
&leqfrac{(H_{m}-h_{m})A_{n}v_{n}+(h_{m}-h)A_{m}v_{m}}{sumlimits_{k=0}^{n}a_{k}v_{k}}
end{align*}