I know there has been many posts about this question here already but I'm not sure how to resolve my dilemma.

***If $ \sum{a_n} $ converges and $\{b_n\}$ is bounded and monotonic, prove that $\sum{a_nb_n}$ converges. ***

This is my work thus far: I have $\sum_na_n$ is convergent which implies for all $\varepsilon > 0$ there exists $N$ s.t.

$\left\vert \sum_{k=n}^ma_k\right\vert =|s_m-s_n|\leq \varepsilon$ for $m \geq n\geq N$ where $\{s_n\}$ is a Cauchy sequence.

$\{b_n\}$ is bounded so there exists $M\in \mathbb{N}$ s.t. $\vert b_n \vert \leq M.$

Then we have $\vert a_nb_n\vert=|a_n|b_n|\leq |a_n|M.$

I know $\left\vert \sum_{n}a_k\right\vert\leq \sum_n|a_n|$ in general but I don't have that $\sum_na_n$ converges absolutely. I haven't learned Dirichlet's Test yet.

What I want to do is show that $|a_nb_n|\leq c_n, n\leq N_0\in \mathbb{N}$ and $\sum c_n$ converges. For now, I think my $c_n=|a_n|M$ but I don't see how I can show $\sum_n |a_n|M=M\sum_n |a_n|$ converges. Any tips please?

  • $\begingroup$ You can not show that $M\sum |a_n|$ converges. It would implies that $\sum a_n$ converges absolutely. But this is not a hypothesis. $\endgroup$
    – PAM1499
    Nov 12, 2020 at 17:05
  • $\begingroup$ Agreed! Do you think the answer I posted below resolves my problem? $\endgroup$ Nov 12, 2020 at 17:11
  • $\begingroup$ I do not understand the following step $|\sum a_kb_k|\leq |\sum a_k M |$ $\endgroup$
    – PAM1499
    Nov 12, 2020 at 17:14
  • $\begingroup$ @PAM1499, would it make sense if I just remove that in the chain of equalities? I wanted the M to be in my equation so I could use it to bound $|\sum a_nb_n|$ to prove convergence. $\endgroup$ Nov 12, 2020 at 19:58
  • $\begingroup$ Just to clarify why what you did is not right define $a_0=-\frac{1}{3}$, $a_n=\frac{(-1)^{n+1}}{2^n}$ for $n \geq 1$ then $\sum a_n=0$. Choose $b_0=0$ and $b_n=1$ for $n \geq 1$. Then $\{b_n\}$ is non-decreasing and bounded by $M=1$. But $|\sum a_nb_n|=\frac{1}{3} >0=|\sum a_n M|$. $\endgroup$
    – PAM1499
    Nov 12, 2020 at 20:12

1 Answer 1


I reworked it; please let me know if this makes sense!!

$\{b_n\}$ is bounded so $|b_n|\leq M$.

$\sum_n a_n$ is convergent so $\forall \varepsilon >0, \exists N$ s.t. $\vert \sum_n^m a_k|\leq \varepsilon/M,\,\,m\leq n \leq N.$

$\vert \sum_{k=n}^m a_kb_k\vert\leq \vert \sum_{k=n}^m a_k M|=|M\sum_{k=n}^m a_k\vert=M\vert\sum_{k=n}^m a_k\vert\leq M\cdot \varepsilon/M=\varepsilon.$


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