Application of Abelian distribution summation method (I.)

1. (and Difference transformation formula) set $m<n$.
$$\sum_{k=m}^{n} (A_{k}-a_{k-1}) B_{k}=a_{n}b_{n}-a_{m-1}b_{m}+\sum_{k=m}^{n-1}a_{k} (b_{k}-b_{k+1}) $$
Proof: Direct calculation can be.
\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. (Division summation method) set $s_{k}=a_{1}+a_{2}+\cdots+a_{k}, (K=1,2,\cdots,n) $. Then
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k} (b_{k}-b_{k+1}) $$
Proof: Supplementary definition $s_{0}=0$, using the conclusion of the first question can be. Make this proposition and the first question equivalent. \\
You may wish to set up a $m<n$, known
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k} (b_{k}-b_{k+1}) $$
$$\sum_{k=1}^{m-1}a_{k}b_{k}=s_{m-1}b_{m-1}+\sum_{k=1}^{m-2}s_{k} (b_{k}-b_{k+1}) $$
Two-type subtraction
$$\sum_{k=m}^{n}a_{k}b_{k}=s_{n}b_{n}-s_{m-1}b_{m-1}+\sum_{k=m-1}^{n-1}s_{k} (b_{k}-b_{k+1}) $$\\
3. Set $s_{n}=a_{1}+a_{2}+\cdots+a_{n}\to s (n\to \infty) $, then
$$\sum_{k=1}^{n}a_{k}b_{k}=sb_{1}+ (S_{n}-s) b_{n}-\sum_{k=1}^{n-1} (S_{k}-s) (B_{k+1}-b_{k}) $$
Proof: By the distribution of the sum of knowledge
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}-\sum_{k=1}^{n-1}s_{k} (B_{k+1}-b_{k}) $$
and
$ $s (b_{n}-b_{1}) =s\sum_{k=1}^{n-1} (B_{k+1}-b_{k}) $$
The two-type subtraction is the conclusion. \\
4. (Abell lemma) if for all $n=1,2,3,\cdots$
$ $b _{1}\geq b_{2}\geq \cdots \geq b_{n}\geq 0$$
$ $m \leq A_{1}+a_{2}+\cdot+a_{n}\leq m$$
Then there are
$ $b _{1}m\geq A_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}\leq b_{1}m$$
Proof: Set $s_{k}=a_{1}+a_{2}+\cdots+a_{k}, (K=1,2,\cdots,n) $. Because $b_{k}\geq 0,b_{k}-b_{k+1}\geq 0$
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k} (B_{k}-b_{k+1}) \leq b_{n}m+m\sum_{k=1}^{n-1} (b_{k}-b _{K+1}) =b_{1}m$$
The left inequality proves similar
$$\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k} (B_{k}-b_{k+1}) \geq b_{n}m+m\sum_{k=1}^{n-1} (b_{k}-b _{K+1}) =b_{1}m$$\\
5. Set $a_{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$ to any real or complex number, and set
$ $A =\max \{|a_{1}|,|a_{1}+a_{2}|,\cdots,|a_{1}+a_{2}+\cdots+a_{n}|\}$$
The
$$\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\}$$
Proof: Use absolute value inequality to indent.
$$\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\}$$
6. (Kronecker) Set $\varphi (n) >0,\varphi (n) \uparrow \infty (n \to \infty) $. Also set $\sum_{n=1}^{\infty}a_{n}$ convergence. The
$$\sum_{k=1}^{n}a_{k}\varphi (k) =o (\varphi (n)), (n\to \infty) $$
Proof: Set $s_{n}\to s (n\to \infty) $ then for any $\varepsilon>0$ existence $m>0,n>m$
$$|s_{n}-s|<\varepsilon$$
By the Division summation method (repeat the third question of the second question) step
\begin{align*}
\sum_{k=1}^{n}a_{k}b_{k}&=s\varphi (1) + (s_{n}-s) \varphi (n)-\sum_{k=1}^{n-1} (S_{k}-s) (\varphi (k+1)-\varphi (k ))\\
&=o (1) +o (\varphi (n))-\sum_{k=1}^{m} (S_{k}-s) (\varphi (k+1)-\varphi (k))-\sum_{k=m+1}^{n-1} (S_{k}-s) (\varphi ( k+1)-\varphi (k)) \ \
&\leq O (1) +o (\varphi (n)) +\varepsilon (\varphi (n)-\varphi (m+1))
\end{align*}
Since $\varphi (n) \to \infty$, know $\sum_{k=1}^{n}a_{k}\varphi (k) =o (\varphi (n)), (n\to \infty) $\\
7. Set $\varphi (n) \downarrow 0 (n \to \infty) $, and $\sum_{n=1}^{\infty}a_{n}\varphi (n) $ for convergence, then
$$\lim_{n\to\infty} (A_{1}+a_{2}+\cdots+a_{n}) \varphi (n) =0$$
Proof: Consider $a_{n}\varphi (n) $ as the $a_{n}$ in Proposition 6, and $\varphi^{-1} (n) $ as the $\varphi (n) $ in Proposition 6, the proposition is proven. \\
\textbf{another certificate}: Direct use of the Abbe lemma proved. Any given $\varepsilon>0$, known by the Cauchy convergence Criterion, exists $n$, $n >n$
$$-\frac{\varepsilon}{2}<a_{n}\varphi (n) +a_{n+1}\varphi{n+1}+\cdots+a_{n}\varphi (n) <\frac{\varepsilon}{2 }$$
and a monotonically decreasing nonnegative column.
$$\varphi^{-1} (N) \geq \varphi^{-1} (n-1) \geq \cdots\geq\varphi^{-1} (N) >0$$
Redefine the sequence, using the Abell lemma Proposition 4, which is
$$|a_{n}+a_{n+1}+\cdot+a_{n}|<\varphi^{-1} (N) \frac{\varepsilon}{2}$$
That
$$| (A_{n}+a_{n+1}+\cdot+a_{n}) \varphi (n) |<\varphi^{-1} (N) \frac{\varepsilon}{2}$$
Because $\varphi (n) \to 0,n\to \infty$
$$\limsup| (A_{1}+\cdots+a_{n-1}+a_{n}+\cdots+a_{n}) \varphi (N) |\leq 0+\frac{\varepsilon}{2}$$
The proposition is to be proven. \ \
8. (Dirichlet) Set $\sigma>0$, the following Dirichlet series
$ $a _{1}\cdot 1^{-\sigma}+a_{2}\cdot 2^{-\sigma}+a_{2}\cdot 3^{-\sigma}+\cdots+a_{n}\cdot n^{-\sigma}+\cdots$$
When convergent, there must be
$$\lim_{n\to\infty} (A_{1}+a_{2}+\cdots+a_{n}) n^{-\sigma}=0$$
Proof: This proposition is a special case of Proposition 7. \\
9. Set $\{z_{n}\}_{1}^{\infty}$ to any one complex column and
$$\sum_{n=1}^{\infty}|z_{n+1}^{-1}-z_{n}^{-1}|=\infty$$
and set the series $\sum_{n=1}^{\infty}a_{n}z_{n}$ for convergence, then there must be
$$\lim_{n\to\infty}\left (\sum_{n=1}^{n}a_{n}\right) \left (\sum_{n=1}^{n}|z_{n+1}^{-1}-z_{n}^{-1}|\right) ^{-1}=0 $$
Proof: This type of topic is the $|\sum a_{n}|$ estimate, set $s_{n}=\sum_{k=1}^{n}a_{k}z_{k},s_{n}\to s (n\to \infty) $, there is an estimate
\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*}
Known by $\lim_{n\to \infty}s_{n}-s=0$, existence $m$, when $n>m$, $|s_{n}-s|<\varepsilon$
Therefore, the right side of the above equation can be shrunk
\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*}
So
\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*}
Thus
$$\limsup \frac{\left|\sum_{n=1}^{n}a_{n}\right|} {\sum_{n=1}^{n}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|} \leq 2\varepsilon$$
Since $\varepsilon$ is an arbitrary number
$$\limsup \frac{\left|\sum_{n=1}^{n}a_{n}\right|} {\sum_{n=1}^{n}\left|z_{n+1}^{-1}-z_{n}^{-1}\right|} =0$$
Proof Method: Abelian Division summation method + piecewise estimate + upper limit. \\
10. When $k=1,2,3,\cdots$ is established, there is
$ $b _{k}\geq b_{k+1},\frac{1}{2} (b_{k}+b_{k+1}) \geq b_{k+1}$$
And
$ $m \leq S_{1}+s_{2}+\cdots+s_{k}\leq m$$
Where $s_{k}=a_{1}+a_{2}+\cdots+a_{k}$. The following inequalities are established
$ $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}$$
Proof: Set
$ $M _{n}=\sum_{k=1}^{n}s_{k},\, \tilde{b}_{k}=b_{k}-b_{k+1}$$
by $b_{k}+b_{k+2}\geq 2b_{k+1}$ $\tilde{b}_{k}\geq \tilde{b}_{k+1}$. Summation Formula by distribution
\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*}
The left inequality proves similar. \\
11. Set $n$ as a fixed large integer, $a _{1},a_{2},\cdots,a_{n},b_{1},b_{2},\cdots,b_{n}$ for any two sets of constants. This definition $b_{k}=0 (k>n) $ and
$$\delta^{m}b_{k}=\delta^{m-1}b_{k+1}-\delta^{m-1}b_{k},\, \delta b_{k}=b_{k+1}-b_{k}$$
$ $s _{k}^{(m)}=\sum_{\nu=1}^{k}s_{\nu}^{(m-1)},s_{k}^{(1)}=a_{1}+a_{2}+\cdots+a_{k}$$
The following identities are established:
$$\sum_{k=1}^{n}a_{k}b_{k}= ( -1) ^{m}\sum_{k=1}^{n}s_{k}^{(m)}\delta^{m}b_{k}$$
Proof: The repeated use of the distribution summation formula can be
\begin{align*}
\sum_{k=1}^{n}a_{k}b_{k}&=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}\left (-\delta b_{k}\right) \ \
&=s_{n}b_{n}+s_{n-1}^{(2)}\left (-\delta b_{n-1}\right) +\sum_{k=1}^{n-2}s_{k}^{2}\left (\Delta^{2} b_{k}\right )\\
&=s_{n}^{(1)}b_{n}+s_{n-1}^{(2)}\left (-\delta b_{n-1}\right) +s_{n-2}^{(3)}\left (\delta^{2} b_{N-2}\right) +\ cdots+ ( -1) ^{n-1}s_{1}^{(N)}\delta^{n-1}b_{1}\\
&=\sum_{k=1}^{n} ( -1) ^{k-1}s_{n-k+1}^{(k)}\delta^{k-1}b_{n-k+1}
\end{align*}
12. Set $a_{k}>0,b_{k}>0,$ and $\{v_{k}\}$ as a monotone descending positive sequence. Also set
$ $H =\max\left (\frac{b_{0}}{a_{0}},\frac{b_{1}}{a_{1}},\cdots,\frac{b_{n}}{a_{n}}\right), \,\,h=\min\left (\frac{ B_{0}}{a_{0}},\frac{b_{1}}{a_{1}},\cdots,\frac{b_{n}}{a_{n}}\right) $$
$ $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) $$
Here $a_{k}=a_{1}+a_{2}+\cdots+a_{k},\, b_{k}=b_{1}+b_{2}+\cdots+b_{k}.$ the following inequalities are established:
$ $h _{m}+ (h-h_{m}) \frac{\sum_{k=0}^{m}a_{k}v_{k}}{\sum_{k=0}^{n}a_{k}v_{k}}\leq \frac{\sum_{k=0}^{n}b_{k}v_{k}}{ \sum_{k=0}^{n}a_{k}v_{k}}\leq h_{m}+ (h-h_{m}) \frac{\sum_{k=0}^{m}a_{k}v_{k}}{\sum_{k=0}^{n}a_{k}v_{k}}$$
Proof: Partial summation method + piecewise estimation, only the left inequality of right inequality proves similar.
\begin{align*}
\frac{\sum_{k=0}^{n}b_{k}v_{k}}{\sum_{k=0}^{n}a_{k}v_{k}}-h_{m}&=\frac{\sum_{k=0}^{m-1}b_{k} (V_{k}-v_{k+1} ) +\sum_{k=m}^{n}b_{k} (V_{k}-v_{k+1}) +b_{n}v_{n}}{\sum_{k=0}^{n}a_{k} (v_{k}-v_{k+1}) +a_{n}v_{n}}-h_{m}\\
&\leq \frac{h\sum_{k=0}^{m-1}a_{k} (V_{k}-v_{k+1}) +h_{m}\sum_{k=m}^{n}a_{k} (v_{k}-v_{k+1}) +H_{m}A_{n}v_{n}}{ \sum_{k=0}^{n}a_{k} (v_{k}-v_{k+1}) +a_{n}v_{n}}-h_{m}\\
&=\frac{(H-h_{m}) \sum_{k=0}^{m-1}a_{k} (V_{k}-v_{k+1})}{\sum_{k=0}^{n}a_{k} (v_{k}-v_{k+1}) +A_{n}v_{n}}\\
&\leq (H-h_{m}) \frac{\sum_{k=0}^{m}a_{k}v_{k}}{\sum_{k=0}^{n}a_{k}v_{k}}
\end{align*}
13. Retain all assumptions of Proposition 12, but change the $\{v_{n}\}$ to a monotonically ascending sequence. Then there are
$ $H _{m}-\frac{(h_{m}-h_{m}) a_{n}v_{n}+ (H-h_{m}) A_{m}v_{m}}{\sum_{k=0}^{n}a_{k}v_{k}}\leq \frac{\sum_{k=0}^{n}b_ {K}v_{k}}{\sum_{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_{k=0}^{n} a_{k}v_{k}}$$
Proof: Partial summation method + piecewise estimation, only the left inequality of right inequality proves similar.
\begin{align*}
\frac{\sum_{k=0}^{n}b_{k}v_{k}}{\sum_{k=0}^{n}a_{k}v_{k}}-h_{m}&=\frac{\sum_{k=0}^{m-1}b_{k} (V_{k}-v_{k+1} ) +\sum_{k=m}^{n}b_{k} (V_{k}-v_{k+1}) +b_{n}v_{n}}{\sum_{k=0}^{n}a_{k} (v_{k}-v_{k+1}) +a_{n}v_{n}}-h_{m}\\
&=\frac{\sum_{k=0}^{m-1} (B_{k}-h_{m}a_{k}) (V_{k}-v_{k+1}) +\sum_{k=m}^{n} (B_{k}-h_{m}a_{k}) (v_{k}-v_{k+1}) + (B_{n}-h_{m}a_{n}) V_{n}}{\sum_{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_{k=0}^{n}a_{k}v_{k}}
\end{align*}

Application of the summation method of abelian distributions (i)