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

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Homework Help Overview

The problem involves proving the convergence of the sequence {bn}n=1∞ given that both {an}n=1∞ and {an + bn}n=1∞ converge. The context is within the study of sequences and their limits in mathematical analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to outline a proof using the definitions of convergence and inequalities. Some participants question the validity of the steps taken, particularly in the manipulation of inequalities and the definition of N2. Others suggest considering the algebra of limits as an alternative approach.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original proof attempt. There are indications of confusion and a lack of clarity in certain steps, prompting the original poster to reflect on their understanding of the material. No consensus has been reached yet.

Contextual Notes

Participants note potential gaps in the original proof, such as the definition of N2 and the handling of inequalities. The original poster expresses a desire to revisit the material for better comprehension.

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Homework Statement



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.

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 {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.
 
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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
 
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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.
 


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]
 


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.
 

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