Real Analysis: Product of sequences diverges

[tarheelborn]

Homework Statement

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

Homework Equations

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.

 

[JG89]

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

 

[tarheelborn]

Sorry... b_n converges to M > 0.

 

[lanedance]

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

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

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

so think about how to pick N based on the behaviour of b_n & a_n

 

[tarheelborn]

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