跟柯西积有关的一个极限证明

╰☆ヾo.海x

请教各位大侠。。这个咋整啊...这是在我们讲柯西积的证明的时候提到的一个lemma

爪机专用

看不懂

icesheep

提示：\[\sum\limits_{k = 0}^n {\left| {{b_{n - k}}} \right|}  = {B_n} \to B\]
提示二：\[B\sum\limits_{n = N}^\infty  {\left| {{a_n}} \right|}  - \sum\limits_{n = N}^\infty  {\left( {\left| {{a_n}} \right|{B_n}} \right)}  = \sum\limits_{n = N}^\infty  {\left| {{a_n}} \right|\left( {B - {B_n}} \right)}  \to 0\]

hbghlyj

Rod Haggarty - Fundamentals of Mathematical Analysis-Addison-Wesley UK (1993) page 118, Chapter 4 Series, § 4.2.9
If $\sum_{r=1}^{\infty} a_r$ and $\sum_{r=1}^{\infty} b_r$ are absolutely convergent series and
\[
c_n=a_1 b_n+a_2 b_{n-1}+\ldots+a_n b_1
\]
then $\sum_{r=1}^{\infty} c_r$ is absolutely convergent and
\[
\sum_{r=1}^{\infty} c_r=\left(\sum_{r=1}^{\infty} a_r\right)\left(\sum_{r=1}^{\infty} b_r\right)
\]
Proof
Suppose first that $\sum_{r=1}^{\infty} a_r$ and $\sum_{r=1}^{\infty} b_r$ are positive-term series and consider the array
\[
\begin{array}{llll}
a_1 b_1 & a_1 b_2 & a_1 b_3 & \ldots \\
a_2 b_1 & a_2 b_2 & a_2 b_3 & \ldots \\
a_3 b_1 & a_3 b_2 & a_3 b_3 & \ldots
\end{array}
\]
If $w_n$ is the sum of the terms in the array that lie in the $n \times n$ square with $a_1 b_1$ at one corner then $w_n=s_n t_n$, where $s_n$ and $t_n$ are the $n$th partial sums of $\sum_{r=1}^{\infty} a_r$ and $\sum_{r=1}^{\infty} b_r$ respectively. Hence
\[
w_n \rightarrow\left(\sum_{r=1}^{\infty} a_r\right)\left(\sum_{r=1}^{\infty} b_r\right) \text { as } n \rightarrow \infty
\]
Now $\sum_{r=1}^{\infty} c_r$ is the sum of the terms in the array summed “by diagonals” and so if $u_n$ is the $n$th partial sum of $\sum_{r=1}^{\infty} c_r$ then
\[
w_{[n / 2]} \leqslant u_n \leqslant w_n
\]
where $[x]$ denotes the largest integer not exceeding $x$ (see Example 11 of Section 5.1). By the sandwich rule for sequences (3.1.3),
\[
u_n \rightarrow\left(\sum_{r=1}^{\infty} a_r\right)\left(\sum_{r=1}^{\infty} b_r\right) \quad \text { as } n \rightarrow \infty
\]
as required.
For the general case the above argument can be applied to $\sum_{r=1}^{\infty}\left|a_r\right|$ and $\sum_{r=1}^{\infty} \abs{ b_r}$ to deduce that $\sum_{r=1}^{\infty} c_r$ is absolutely convergent. Since $\sum_{r=1}^{\infty} a_r$ and $\sum_{r=1}^{\infty} b_r$ are linear combinations of $\sum_{r=1}^{\infty} a_r^{+}, \sum_{r=1}^{\infty} a_r^{-}$, and $\sum_{r=1}^{\infty} b_r^{+}, \sum_{r=1}^{\infty} b_r^{-}$ respectively, it is readily established that
$$\sum_{r=1}^{\infty} c_{r}=\left(\sum_{r=1}^{\infty} a_{r}\right)\left(\sum_{r=1}^{\infty} b_{r}\right)$$by 4.1.3 and 4.1.4.    $_\square$