Register to reply

Sequence: product of sequences diverges

by tarheelborn
Tags: diverges, product, sequence, sequences
Share this thread:
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
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
Mark44
#2
Oct13-09, 07:56 PM
Mentor
P: 21,216
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: 21,216
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: 21,216
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: 21,216
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.


Register to reply

Related Discussions
Real Analysis: Product of sequences diverges Calculus & Beyond Homework 4
Proof: c * divergent sequence diverges Calculus & Beyond Homework 1
Finding an oscillating sequence that diverges and whose limit is zero. Calculus & Beyond Homework 2
Proving a sequence diverges with limited information Calculus 1
What is the formal proof that a sequence diverges .. Calculus & Beyond Homework 2