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 $A \le |\sum_{j=m+1}^{n}x_j|$?
