Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.