Dirichlet's Test Remark in Apostol Dirichlet's Test is theorem $10.17$ in Apostol's Calculus Vol. $1$.
The theorem itself says that if the partial sums of $\{a_n\}$ (can be complex numbers, not just reals) form a bounded sequence and $\{b_n\}$ is a (monotone?) decreasing function converging to $0$, then $\sum a_n b_n$ converges.
The part of the proof I am stuck on says that, letting $A_n=\sum_{k=1}^{n} a_k$
"The series $\sum (b_k - b_{k+1})$ is a convergent telescoping series which dominates $\sum A_k(b_k - b_{k+1})$.  This implies absolute convergence..."
How does this imply absolute convergence?  Does it have to do with the fact that $\{b_n\}$ is decreasing?  By decreasing, should I automatically think monotone?
 A: Note that the partial sums of $\{a_n\}$ are bounded means that $\lvert A_k \rvert \leq M$ for all $k$ and some $M > 0$. Hence, we have that
\begin{align}
\left \lvert \sum_{k \leq n} A_k(b_k - b_{k+1}) \right \rvert & \leq \sum_{k \leq n} \left(\left \lvert A_k(b_k - b_{k+1}) \right \rvert \right) & (\because \text{By triangle inequality})\\
&= \sum_{k \leq n} \left \lvert A_k \right \rvert \left \lvert (b_k - b_{k+1}) \right \rvert & \because \lvert z_1 z_2 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert\\
& \leq \sum_{k \leq n} M \lvert(b_k - b_{k+1}) \rvert & (\because A_k \text{ is bounded by }M)\\
& = M \sum_{k \leq n} (b_k - b_{k+1}) & (\because \{b_n\}\text{ form a decreasing sequence})\\
& = M (b_1 - b_{n+1}) & (\because \text{By telescoping})\\
& \leq Mb_1 & (\because b_n \downarrow 0 \implies b_{n+1} \geq 0)
\end{align}
Hence, $\displaystyle \sum_{k \leq n} A_k(b_k - b_{k+1})$ converges absolutely.
A: The could be a typo in the book, where the absolute value is missing, because $A_k$ could be complex, so the telescopic series actually dominates $\sum |A_k (b_k - b_{k+1})|$. Otherwise, I don't see how the relationship even makes sense, because $\le$ is not defined between complex numbers. The rest of the proof is the same as in the book.
