Is the Product of a Convergent Series and a Bounded Sequence Also Convergent?

[squaremeplz]

Homework Statement

If the sequence of partial sums of |a_n| is convergent and b_n is bounded, prove that the sequence of partial sums of the product (a_n)(b_n) is also convergent.

Homework Equations

Cauchy sequences and bounded sequences

The Attempt at a Solution

I wrote the following

for n,m > N_1 and e > 0

|a_n - a_m | < e

which proves that a_n is a cauchy sequence, for every convergent sequence is a cauchy sequence.

For b_n, we assume e = 1, and for n, m > N_2

we have

|b_n - b_m | < 1

then for N = max{N_1, N_2}

|a_n * b_n - a_m * b_m | < e + 1

 

[HallsofIvy]

Homework Statement

If the sequence of partial sums of |a_n| is convergent and b_n is bounded, prove that the sequence of partial sums of the product (a_n)(b_n) is also convergent.

Homework Equations

Cauchy sequences and bounded sequences

The Attempt at a Solution

I wrote the following

for n,m > N_1 and e > 0

|a_n - a_m | < e

which proves that a_n is a cauchy sequence, for every convergent sequence is a cauchy sequence.

Well, no. "|a_n- a_m|< e" doesn't prove that a_n is a Cauchy sequence. The fact that every convergent sequence is a Cauchy sequence proves that |a_n- a_m|< e.
Also, the wording "for n,m> N_1 and e> 0" is incorrect. That implies that you can find m,m> N_1 such |a_n- a_m|< e for all e> 0 which is not true. Given e> 0 first, you can find N_1 such that that is true.

For b_n, we assume e = 1, and for n, m > N_2

we have

|b_n - b_m | < 1

This is certainly NOT true. Your only condition on {b_n} is that it is bounded, not that it is convergent. For example, {b_n}= {(-1)^n} is bounded and this is not true

then for N = max{N_1, N_2}

|a_n * b_n - a_m * b_m | < e + 1

What does "bounded" mean?