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?

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This is closely related to partial summation, which is a powerful technique. – AD. Jun 19 '12 at 7:12
Actually, thinking about it like integration by parts helps a lot. Thanks. – Andrew Salmon Jun 19 '12 at 7:17

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.
So by "decreasing sequence," they mean "monotone decreasing" then. So that means that $\{b_n\}$ is positive? – Andrew Salmon Jun 19 '12 at 6:42
@AndrewSalmon Decreasing sequence means $b_n \geq b_{n+1}$. It is a good exercise to show that if a sequence is decreasing and converges to $0$, then $b_n \geq 0$. – user17762 Jun 19 '12 at 6:44