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Sequence: product of sequences diverges

by tarheelborn
Tags: diverges, product, sequence, sequences
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tarheelborn
#1
Oct13-09, 07:48 PM
P: 123
1. The problem statement, all variables and given/known data

Sorry, I posted this earlier but I had an error in my problem statement; please advise. Thank you.

If a_n diverges to +inf, b_n converge to M>0; prove a_n*b_n diverges to +inf

2. Relevant equations



3. The attempt at a solution

My attempt follows: I seem to have trouble getting things in the right order, so I am trying to work on my technique, with your help. Also, I am afraid I may have omitted reference to some theorem that I am taking for granted, which is another of my bad habits. Please review for me and advise as appropriate. I am determined to conquer this subject! Thanks.

Let M, e > 0, M, e \in R. By definition of a limit of a sequence, we can choose N_b such that |b_n - M|< e, n >= N. Then -e < b_n - M < e, so M - e < b_n < M + e. So b_n > M - e. We can then choose N_a such that a_n >= M/(M-e), n >= N. Let N = max (N_b, N_a). Then a_n*b_n >= M/(M-e), n >= N. Thus, {a_n*b_n} diverges to + infinity.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Mark44
#2
Oct13-09, 07:56 PM
Mentor
P: 21,258
When you show that a sequence diverges to infinity, you don't use epsilon; you show that starting with some index in the sequence, all elements are larger than some (large) number M.

The definition for such a sequence goes something like this:
For all numbers M > 0, there exists an index N such that, if n >= N, a_n > M.

You already know that a_n gets large without bound. What can you say about a_n*b_n? Make sure that you don't use M for both a_n and a_n*b_n.
tarheelborn
#3
Oct13-09, 08:02 PM
P: 123
It seems as if I should be able to say a_n*b_n gets (large without bound * L), where L = lim b_n, but I don't know how to say that. Instinctively I know that this happens.

tarheelborn
#4
Oct13-09, 08:12 PM
P: 123
Sequence: product of sequences diverges

What if I say:

--------------------------------------------------------------------------------

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

If a_n diverges to +inf, b_n converge to 0; prove a_n*b_n diverges to +inf

2. Relevant equations



3. The attempt at a solution

My attempt follows: I seem to have trouble getting things in the right order, so I am trying to work on my technique, with your help. Also, I am afraid I may have omitted reference to some theorem that I am taking for granted, which is another of my bad habits. Please review for me and advise as appropriate. I am determined to conquer this subject! Thanks.

Let L, M, e > 0, L, M, e \in R. By definition of a limit of a sequence, we can choose N_b such that |b_n - L|< e, n >= N. Then -e < b_n - L < e, so L - e < b_n < L + e. So b_n > L - e. We can then choose N_a such that a_n >= M/(L-e), n >= N. Let N = max (N_b, N_a). Then a_n*b_n >= [M/(L-e)]*L-e, which = M, n >= N. Thus, {a_n*b_n} diverges to + infinity.
Mark44
#5
Oct13-09, 08:26 PM
Mentor
P: 21,258
Quote Quote by tarheelborn View Post
What if I say:

--------------------------------------------------------------------------------

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

If a_n diverges to +inf, b_n converge to 0; prove a_n*b_n diverges to +inf
You've changed the problem on me! In the first post in this thread, b_n converged to a positive number.

Which is it?
tarheelborn
#6
Oct13-09, 08:28 PM
P: 123
It does converge to a positive number. SORRY. I posted this twice, once had an error and I seem to be picking up the wrong one somehow. The problem is: if a_n diverges to +infinity and b_n converges to M>0, prove a_n*b_n converges to +infinity.
Mark44
#7
Oct13-09, 08:36 PM
Mentor
P: 21,258
OK, that's better. Now go back and read what I wrote in post #2. Epsilon doesn't enter into things at all when your trying to show that a sequence diverges to infinity. You can't get within epsilon of infinity, and that seems to be what you're trying to do.
tarheelborn
#8
Oct13-09, 08:44 PM
P: 123
I understand that I need to say that a_n*b_n > M, therefore a_n*b_n diverges. But don't I have to initially deal with epsilon in the sense that the initial sequence b_n DID converge to a limit? I was trying to get that limit cancelled out so that the "result" was that the product is greater than my M. I am not trying to be stubborn; I just don't get it. The only thing I get in class is the proof copied from my book onto the chalkboard. It just hasn't sunk in yet.
Mark44
#9
Oct13-09, 10:14 PM
Mentor
P: 21,258
Quote Quote by tarheelborn
Let L, M, e > 0, L, M, e \in R. By definition of a limit of a sequence, we can choose N_b such that |b_n - L|< e, n >= N. Then -e < b_n - L < e, so L - e < b_n < L + e. So b_n > L - e. We can then choose N_a such that a_n >= M/(L-e), n >= N. Let N = max (N_b, N_a). Then a_n*b_n >= [M/(L-e)]*L-e, which = M, n >= N. Thus, {a_n*b_n} diverges to + infinity.
On a closer look, and with your later explanation, this looks pretty good. I don't see anything wrong in your argument.

I got confused when you changed the limit of your converging sequence from a positive number to 0, which completely changes the outcome.

One thing I would suggest is to spread things out a bit to improve readability. What you have is very dense, making it more difficult to comprehend.
tarheelborn
#10
Oct14-09, 10:13 AM
P: 123
Thank you so much! I will definitely try to work on making things more readable. I really appreciate your help.


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