converging analysis proof


by dancergirlie
Tags: analysis, converging, proof
dancergirlie
dancergirlie is offline
#1
Sep26-09, 09:27 PM
P: 200
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!
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lanedance
lanedance is offline
#2
Sep26-09, 11:10 PM
HW Helper
P: 3,309
Quote Quote by dancergirlie View Post
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
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|


Quote Quote by dancergirlie View Post

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 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
lanedance
lanedance is offline
#3
Sep26-09, 11:15 PM
HW Helper
P: 3,309
updated above

dancergirlie
dancergirlie is offline
#4
Sep27-09, 11:38 AM
P: 200

converging analysis proof


thanks for the help :)


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