Prove if $\sum\limits_{n=1}^ \infty a_n$ converges, {$b_n$} is bounded & monotone, then $\sum\limits_{n=1}^ \infty a_nb_n$ converges.

Prove that if $\displaystyle \sum_{n=1}^ \infty a_n$ converges, and {$b_n$} is bounded and monotone, then $\displaystyle \sum_{n=1}^ \infty a_nb_n$ converges.

No, $a_n, b_n$ are not necessarily positive numbers.

I've been trying to use the Dirichlet's Test, but I have no way to show that $b_n$ goes to zero. If I switch $a_n$ and $b_n$ in Dirichlet's Test, I can show $a_n$ goes to zero, but then I'm having trouble showing that $\displaystyle \sum_{n=1}^ \infty \left\lvert a_{n+1}-a_{n}\right\rvert$ converges (because $a_n$ isn't necessarily monotone).

• Use abel partial summation. Commented Feb 5, 2016 at 0:31
• I have, that is what allowed me to show that the partial sums of ∑anbn converge. But, I still need to satisfy the other condition(s). It's these other conditions am having difficulty with. Commented Feb 5, 2016 at 0:56
• Since {$b_n$} is bounded and monotone, it has a Max/Min (both cases are analogous). So let $M$ be a lower/upper bound (which is finite) for {$b_n$}, then $M\cdot\sum a_{n}$ converges... shouldn't be hard from there, or I could be missing something? Commented Feb 5, 2016 at 1:25
• @mm8511 With this argument, you don't show convergence of the series, only that it's bounded. Commented Feb 5, 2016 at 1:41

Outline: Indeed, the key is use Dirichlet's test (a.k.a. Abel's summation at its core) as you intended:

\begin{align} \sum_{n=1}^N a_n b_n &= \sum_{n=1}^N (A_n-A_{n-1}) b_n = A_Nb_N + \sum_{n=1}^{N-1} A_n b_n -\sum_{n=1}^{N-1} A_n b_{n+1} \\ &= A_Nb_N + \sum_{n=1}^{N-1} \underbrace{A_n}_{\text{bounded}} \underbrace{(b_n -b_{n+1})}_{\text{constant sign}} \end{align} where $A_n \stackrel{\rm def}{=} \sum_{n=1}^{N} a_n$ and $A_0=0$.

Now, this does not quite work: the issue boils down to the fact that at the end of the day you cannot rely on $b_n\xrightarrow[n\to\infty]{} 0$; since indeed it is not necessarily true. But you need this for the argument to go through.

To circumvent that, observe that $(b_n)_n$ is a bounded monotone sequence, and therefore is convergent. Let $\ell\in\mathbb{R}$ be its limit.

You can now define the sequence $(b^\prime_n)_n$ by $b^\prime_n \stackrel{\rm def}{=} b_n-\ell$. This is a monotone bounded sequence converging to $0$, and $$\sum_{n=1}^N a_n b^\prime_n = \sum_{n=1}^N a_n b_n - \ell \sum_{n=1}^N a_n.$$ The second term is a convergent series by assumption on $(a_n)_n$, so showing convergence of the series $\sum a_n b_n$ is equivalent to showing convergence of the series $\sum a_n b^\prime_n$. Apply your idea (Abel's summation) on the latter.

• That's definitely what I was looking for. Thank you so much. Commented Feb 5, 2016 at 2:15
• Glad this was helpful... that's a nice question. Commented Feb 5, 2016 at 2:16
• Can I conclude that OP's statement is a corollary of Dirichlet's test? Commented Jun 18, 2019 at 6:02
• Sure, you can phrase it that way. @ChoMedit. Commented Jun 18, 2019 at 13:54