# Thread: Multiplication of convergent series

1. ## Multiplication of convergent series

Given that $\displaystyle \sum{a_n}$ converges and $\displaystyle [b_n]$ is monotonic and bounded, prove $\displaystyle \sum{a_nb_n}$ converges.

So I know $\displaystyle a_n{\rightarrow{0}}$ and $\displaystyle b_n$ converges, so $\displaystyle a_nb_n{\rightarrow{0}}$
Where do I go from here?

2. ## Re: Multiplication of convergent series

Originally Posted by I-Think
Given that $\displaystyle \sum{a_n}$ converges and $\displaystyle [b_n]$ is monotonic and bounded, prove $\displaystyle \sum{a_nb_n}$ converges.

So I know $\displaystyle a_n{\rightarrow{0}}$ and $\displaystyle b_n$ converges, so $\displaystyle a_nb_n{\rightarrow{0}}$
Where do I go from here?
PlanetMath: proof of Abel's test for convergence

3. ## Re: Multiplication of convergent series

Let $\displaystyle S_n:=\sum_{k=1}^na_kb_k$, $\displaystyle s_k:=\sum_{j=1}^ka_k$ and $\displaystyle M:=\sup_{j\in\mathbb N}|s_j|<+\infty$ since the sequence $\displaystyle \{s_j\}$ is bounded. We have for $\displaystyle n\geq 1$:
\displaystyle \begin{align*}S_n&=\sum_{k=1}^n(s_k-s_{k-1})b_k\\&=\sum_{j=1}^ns_jb_j-\sum_{j=0}^{n-1}s_jb_{j+1}\\&=\sum_{j=1}^ns_j(b_j-b_{j+1})+s_nb_n.\end{align*}
We have to show that the sequence $\displaystyle \{S_n\}$ is a Cauchy sequence. To see that, write for $\displaystyle m,n\geq 1$:
\displaystyle \begin{align*}|S_{m+n}-S_n|&=\left|\sum_{j=m+1}^{m+n}s_j(b_j-_{j+1}) +s_{m+n}b_{m+n}-s_{m+1}b_{m+1}\right| \\&\leq\left|\sum_{j=m+1}^{m+n}s_j(b_j-_{j+1})\right|+\left|a_{m+n}b_{m+n}-a_{m+1}b_{m+1}\right|\\&\leq M|b_{m+1}-b_{m+n}|+\left|a_{m+n}b_{m+n}-a_{m+1}b_{m+1}\right|.\end{align*}