# Converging analysis proof

1. Sep 26, 2009

### dancergirlie

1. The problem statement, all variables and given/known data

Assume that (an) is a bounded (but not necessarily convergent) sequence, and that the
sequence (bn) converges to 0. Prove that the sequence (anbn) converges to zero.

2. Relevant equations

3. The attempt at a solution

Assume that an is a bounded sequence and bn converges to 0.

That means for all n in N, there exists a M >0 so that
|an|<=M
Since bn converges, that means that it must be bounded as well. Which means for all n in N there exists a P>0 so that
|bn|<=P

since |an|<=M and |bn|<=P that means for all n in N:
|an||bn|<= MP which is equivalent to |anbn|<=MP
where MP>0 since M>0 and P>0. Hence (anbn) is bounded

Since bn converges to 0 that means for e>0 there exists an N in N so that for n>=N
|bn-0|<e
which is equivalent to -e<bn<e

This is where I get stuck. Do I just multiply the inequality by an? cause then I'd have
-e(an)<bnan<e(an)
which would be equivalent to |anbn|<e2 if I let e2=e(an) which would mean that anbn converges to zero as well. But I don't know if I can multiply the sequence by it though....

Any help would be great!

2. Sep 26, 2009

### lanedance

an bounded look alright
As bn is convergent, I would say for any P>0, there exists N such that for all n>N then
|bn-0|< |bn|

i think you were almost there...

now what you need to show to prove an.bn converges to zero, is that for any e>0 you can choose N, such that for all n>N you have
|an.bn|<e

as you know an<=M for all n, then
|an.bn|<=|M.bn|

so now you just need to show you can choose N such that for all n>N
|bn|<=e/|M|
and i think you're there

3. Sep 26, 2009

### lanedance

updated above

4. Sep 27, 2009

### dancergirlie

thanks for the help :)