Convergence of Bounded Sequences: Proving the Convergence of (anbn) to Zero

[dancergirlie]

Homework Statement

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.

Homework Equations

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!

 

[lanedance]

Homework Statement

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.

Homework Equations

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

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

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!

i think you were almost there...

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

as you know a_n<=M for all n, then
|a_n.b_n|<=|M.b_n|

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

 

[lanedance]

updated above

 

[dancergirlie]

thanks for the help :)