Multiplication of convergent series

**I-Think** · August 21st 2011, 09:34 AM

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

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

**Also sprach Zarathustra** · August 21st 2011, 09:49 AM

Originally Posted by I-Think

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

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

PlanetMath: proof of Abel's test for convergence

**girdav** · August 21st 2011, 10:39 AM

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

Thread: Multiplication of convergent series

LinkBack

Thread Tools

Display

Multiplication of convergent series

Re: Multiplication of convergent series

Re: Multiplication of convergent series

Similar Math Help Forum Discussions

How can you tell if a series is absolutely convergent or conditionally convergent?

Determine whether the series is absolutely convergent, conditionally convergent ....

Does the subtraction of 2 diverging Series mean the resulting Series is convergent?

determine if series is absolutely convergent conditionally convergent or divergent

Series: Absolutely convergent, con convergent or divergent

Search Tags