Proof That {bn}n=1∞ Converges When {an}n=1∞ and {an + bn }n=1∞ Converge

[Battlemage!]

Homework Statement

Suppose {a_n}_n=1^∞ and {b_n}_n=1^∞ are sequences such that {a_n}_n=1^∞ and {a_n + b_n }_n=1^∞ converge.

Prove that {b_n}_n=1^∞ converges.

Homework Equations

The definition of convergence.

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 {a_n}_n=1^∞ converges to A (by hypothesis).

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

2. Assume that {a_n + b_n }_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 | (a_n + b_n) - (A + B) | < ε​

3. | (a_n + b_n) - (A + B) | = | (a_n - A) + (b_n - B) | < ε

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

| (a_n - A) - (a_n - A) + (b_n - B) | < ε - ε/2

| (b_n - B) | < ε/2

if n ≥ N₂ for some positive integer N₂.​

5. But this is the definition of convergence, therefore {b_n}_n=1^∞ converges (to B). □
Thanks.

 

[Office_Shredder]

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

 

[vela]

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.

 

[hunt_mat]

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

 

[Battlemage!]

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.