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

Battlemage!

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

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.



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 {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 [tex]|a_n-A|[/tex] 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 [tex]\epsilon/2[/tex] 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

[tex]b_n=(a_n+b_n)-a_n[/tex] 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 [tex]a_{n}+b_{n}\rightarrow b}[/tex] and [tex]a_{n}\rightarrow a[/tex] then the sequence [tex]b_{n}=a_{n}+b_{n}-a_{n}\rightarrow b-a[/tex]

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.