Register to reply 
Sequence: product of sequences diverges 
Share this thread: 
#1
Oct1309, 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/(Me), n >= N. Let N = max (N_b, N_a). Then a_n*b_n >= M/(Me), 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 


#2
Oct1309, 07:56 PM

Mentor
P: 21,321

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. 


#3
Oct1309, 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.



#4
Oct1309, 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/(Le), n >= N. Let N = max (N_b, N_a). Then a_n*b_n >= [M/(Le)]*Le, which = M, n >= N. Thus, {a_n*b_n} diverges to + infinity. 


#5
Oct1309, 08:26 PM

Mentor
P: 21,321

Which is it? 


#6
Oct1309, 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.



#7
Oct1309, 08:36 PM

Mentor
P: 21,321

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.



#8
Oct1309, 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.



#9
Oct1309, 10:14 PM

Mentor
P: 21,321

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. 


#10
Oct1409, 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 