I am working from "Understanding Analysis" by Abbot and the following is an exercise that works through the proof of Abel's Test. I reproduce the question and the solution. I am confused at a section of the proof towards the end. Any clarifications would be great.
Abel's Test for convergence states that if the series
$\sum_{n=1}^{\infty}x_n$ converges, and if $(y_n)$ is a sequence
satisfying $y_1 \ge y_2 \ge \cdots \ge 0$, then the series
$\sum_{n=1}^{\infty}x_ny_n$ converges.(a) Assume that $\sum_{n=1}^{\infty}a_n$ has partial sums that are
bounded by a constant $A>0$ and assume $b_1 \ge b_2 \ge \cdots \ge 0$.
Use summation by parts to show that $|\sum_{j=1}^{n} a_jb_j | \le
2Ab_1$.
Let $A>0$ be an upper bound for the partial sums $s_n$ of $\sum_{n=1}^{\infty}a_n$, hence
\begin{align}
|\sum_{j=1}^{n} a_jb_j | &= |s_n b_{n+1} – s_m b_{m+1} + \sum_{j=m+1}^{n} s_j(b_j-b_{j+1})|\le \\
& \le Ab_{n+1} + Ab_{m+1} + A(b_{m+1}-b_{n+1})= \\
& =2Ab_{m+1} \le 2Ab_1
\end{align}
(b) Prove Abel's Test by setting $a_n = x_{m+n}$ and $b_n = y_{m+n}$.
To show that $\sum_{n=1}^{\infty}x_ny_n$ converges, we use the Cauchy Criterion. Let $\epsilon > 0$, we need to show that there exists an $N$ such that if $n > m \ge N$, it follows that $|\sum_{j=m+1}^{n}x_jy_j|<\epsilon$. Let $a_n = x_{m+n}$ and $b_n = y_{m+n}$, then from part (a), we have
\begin{align}
|\sum_{j=m+1}^{n}x_jy_j| = |\sum_{j=1}^{n-m}a_jb_j| \le 2Ab_1,
\end{align}
where $A$ is an upper bound on the partial sums of $\sum_{n=1}^{\infty}a_n=\sum_{j=m+1}^{\infty}x_j$. Since $\sum_{n=1}^{\infty}x_n$ converges, then by the Cauchy Criterion, we can pick $N$ such that $n > m \ge N$ implies $|\sum_{j=m+1}^{n}x_j| < \frac{\epsilon}{2y_1}$.
Up until this point, I clearly understand every step of the proof, but the following is when I get confused:
Looking again at what the constant $A$ represents, it follows that if $n > m \ge N$, then
\begin{align}
A \le |\sum_{j=m+1}^{n}x_j| < \frac{\epsilon}{2y_1}.
\end{align}
How can the above be true? $A$ is the upper bound of the partial sum of the series $\sum_{j=m+1}^{\infty}x_j$, so by definition, we should have $|\sum_{j=m+1}^{n}x_j| \le A$ for all $n > m$. So, why is the above inequality $A \le |\sum_{j=m+1}^{n}x_j|$?
Best Answer
In the proof of Abel's Test, $A$ is a constant relative to $n$ but not constant relative to $m$ and should be written as $A_m,$ and should not be defined as just any upper bound for $\{|\sum_{j=1}^ka_j\}_k ,$ but as $\sup_k |\sum_{j=1}^ka_j|.$ Then $A_m=\sup_k|\sum_{j=1}^k|=\sup_k|\sum_{j=1}^kx_{m+j}|,$ which is as small as we want if we choose $m$ large enough. We obtain $$|\sum_{j=m+1}^nx_jy_j|=|\sum_{j=1}^{n-m}a_jb_j|\leq 2A_mb_1=2A_my_{m+1}$$ I suppose we should also write $a_{n,m}$and $b_{n,m}$ rather than $a_n$ and $b_n$.