# Real Analysis: Product of sequences diverges

1. Oct 13, 2009

### tarheelborn

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 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

2. Oct 13, 2009

### JG89

Counter example: $$a_n = n \left , \left b_n = \frac{1}{n^2}$$

3. Oct 13, 2009

### tarheelborn

Sorry... b_n converges to M > 0.

4. Oct 14, 2009

### lanedance

so you need to show that for any P > 0 you can find N such that anbn > P forall n > N

bn converges to M, so for any e > 0, there exists N1 such that |bn-M|< e for all n>N1

an diverges to infinity, so for any P1 > 0, there exists N2 such that an > P1 for all n > N2

so think about how to pick N based on the behaviour of bn & an

5. Oct 14, 2009

### tarheelborn

Actually, I did get this solved. Thank you so much for your help!