# Homework Help: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} converges

1. Sep 22, 2010

### Battlemage!

1. The problem statement, all variables and given/known data

Suppose {an}n=1 and {bn}n=1 are sequences such that {an}n=1 and {an + bn }n=1 converge.

Prove that {bn}n=1 converges.

2. Relevant equations

The definition of convergence.

3. The attempt at a solution

I am pretty new to mathematics that requires proof, so excuse me if I do something really stupid... but basically, is this a sufficient proof?

1. Assume {an}n=1 converges to A (by hypothesis).

Then for ε/2 > 0 there is a positive integer N1 such that if n ≥ N1, then |an - A| < ε/2.​

2. Assume that {an + bn }n=1 converges to A + B (by hypothesis).

Then for ε > 0 there is a positive integer N = max{N1, N2} such that if n ≥ N, then | (an + bn) - (A + B) | < ε​

3. | (an + bn) - (A + B) | = | (an - A) + (bn - B) | < ε

4. Since by hypothesis |an - A| < ε/2, then

| (an - A) - (an - A) + (bn - B) | < ε - ε/2

| (bn - B) | < ε/2

if n ≥ N2 for some positive integer N2.​

5. But this is the definition of convergence, therefore {bn}n=1 converges (to B).

Thanks.

Last edited: Sep 22, 2010
2. Sep 22, 2010

### Office_Shredder

Staff Emeritus
Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

You never actually say what N2 is.

Also, if a+b<e and a<e/2 that doesn't give you that b<e/2, so I'm not sure where that final inequality comes from. Subtracting $$|a_n-A|$$ from the left hand side does not allow you to just move it inside the absolute value sign; and subtracting |a_n-A| from the right hand side, you can't replace it with $$\epsilon/2$$ and maintain the same inequality, since that makes the right hand side smaller, not larger

As a fast way to see a lot of results like this, once you have the standard summation and multiplication rules, you can use

$$b_n=(a_n+b_n)-a_n$$ and use what you know about the summation of sequences

Last edited: Sep 22, 2010
3. Sep 22, 2010

### vela

Staff Emeritus
Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

There's a problem in step 4 of the attempt. It doesn't follow from

| (an - A) + (bn - B) | < ε

that

| (an - A) - (an - A) + (bn - B) | < ε - ε/2

It's like saying |1-1.9|=0.9 < 1 so |1-1-1.9| < 1-1=0.

4. Sep 22, 2010

### hunt_mat

Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

This looks okay, why not look at the algebra of limits? if $$a_{n}+b_{n}\rightarrow b}$$ and $$a_{n}\rightarrow a$$ then the sequence $$b_{n}=a_{n}+b_{n}-a_{n}\rightarrow b-a$$

5. Sep 23, 2010

### Battlemage!

Re: Convergent sequences: if {an} converges and {an + bn} converges, prove {bn} conve

Thanks everyone. I'm clearly missing something in the understanding of this material, so I'll take what you've said this weekend and dig through the book and see if I can spot the misunderstanding.