Take the 2-minute tour ×
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.

a) Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.

I wanted to use the comparison test to show it's true, but I think I got something mixed up. Here's what I have so far.

$b_n$ is bounded $\Rightarrow \exists c > 0: |b_k| < c, \forall k \in \mathbb N$

$\sum a_nb_n < \sum a_nc < c\sum a_n$. I'm very tempted to use the comparison test here and say, as $\sum a_n$ converges absolutetly, then so does $c\sum a_n$, but for that I needed the inverted relation, right? I would need $\sum a_n > c\sum a_n$, which is not true. However isn't it obvious that something absolutely convergent multiplied by a constant is also absolutetly convergent? Is there a mathematical way to write this? Thanks a lot in advance.

b) Refute with a counter example: if $\sum a_n$ converges and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges.

Is this even possible? I mean, it has to be, but it doesn't make sense to me. If something converges, it means it's bounded, right? And I thought, well, since bounded + bounded = bounded, then bounded*bounded would also get me something bounded again.

Anyway, my idea would be to use for $a_n$ an alternating series that ist only convergent, for example $(-1)^n \frac 1 {n^2}$. But something tells me I'm trying to prove $a_n b_n$ is not absolutely convergent.

Thanks a lot in advance guys!

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

Recall that if $\sum a_n$ is finite then $c\cdot\sum a_n = \sum ca_n$. This is true even if the series is not absolutely convergent.

The reason is simple, $\displaystyle\sum a_n = \lim_{k\to\infty}\sum_{n<k} a_n$, and when multiplying by a constant it can jump into the limit.

Now your reasoning is true. If $b_n\le c$ then $\sum a_nb_n\le \sum ca_n\le c\sum a_n$, and the latter converges.

For the second question, in the first one the assumption was the series is absolutely convergent. Take a sequence which is not of this form.

For example $a_n = \dfrac{(-1)^n}{n}$, and $b_n=(-1)^n$. Now what is $\sum a_nb_n$?

share|improve this answer
    
Thanks for the swift reply asaf! Quick question, my reasoning did not use the "comparison test", right? –  Clash Nov 24 '11 at 17:09
    
@Clash: Yes, but the reasoning that we can replace $b_n$ by their upper bound. –  Asaf Karagila Nov 24 '11 at 17:14
    
and that multiplying something absolutely convergent by a constant does not change it's "status", right? thanks again, i love u –  Clash Nov 24 '11 at 17:17
    
@Clash: You might want to prove that. It follows from $\sum |ca_n| = \sum |c||a_n| = |c|\sum |a_n|<\infty$. –  Asaf Karagila Nov 24 '11 at 17:26
    
perfect, thank you so much! –  Clash Nov 24 '11 at 17:30
add comment

Your Answer

 
discard

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.