Prove Sum of Converging Sequences = L+M, Counter Example

  • Thread starter Design
  • Start date
  • Tags
    Sequences
In summary, we can prove that if two sequences an and bn converge to L and M respectively, then the sum of the sequences converge to L+M by using the triangle inequality and an epsilon/2 argument. This means that for any epsilon greater than 0, there exist positive integers N1 and N2 such that |an - L| < epsilon/2 when N >= N1 and |bn - M| < epsilon/2 when N >= N2. The maximum of N1 and N2 will satisfy both inequalities, proving the statement. Additionally, it is possible to present a counter example to show that the sum of two divergent sequences need not be divergent, by choosing an epsilon value that does not satisfy both inequalities
  • #1
Design
62
0

Homework Statement


Prove that if two sequences an and bn converge to L and M respectively, then the sum of the sequences converge to L+M.
Present a counter example to show the sum of two divergent sequences need not be divergent.

The Attempt at a Solution


We have an -> L and bn -> M we want an + bn -> L+M.

So I have |an - L | < Ε for every n > N
and |bn - L | Ε every n > N

What is my next step?

thank you
 
Last edited:
Physics news on Phys.org
  • #2
Use the triangle inequality and an epsilon/2 argument.
 
  • #3
Is my starting point totally off?
 
  • #4
No, it is quite good. But you may want to modify what you have in order to distinguise the n's for which a_n and b_n can be made smaller than epsilon. It would also be useful to put a_n and b_n in the epsilon/2 neighbourhood of L and M. You will see why later.

|a_n+ b_n -(L+M)|=|a_n-L + b_n -M|

Use the triangle inequality on the above and then select your n_0 and then use the modified part that you already have.
 
  • #5
How can i make an and bn with the E/2 argument, don't understand that part. I know we are trying to prove that statement.
 
  • #6
As ╔(σ_σ)╝ noted, it's only a point of rearranging terms, making use of the triangle inequality and chosing some ε > 0, and some ε/2 > 0. I.e. if an and bn coverge, then for ε/2 (whatever ε is) there exist some positive integers N1 and N2 such that |an - L| < ε/2 whenever N >= N1 and |bn - M| < ε/2 whenever N >= N2. For which integers will both inequalities be satisfied?
 
  • #7
maximum of n1 n2?
 
  • #8
Yes :-).

Show us what you have done.
 

Related to Prove Sum of Converging Sequences = L+M, Counter Example

What is meant by "Prove Sum of Converging Sequences = L+M, Counter Example"?

This statement refers to a mathematical concept where the sum of two converging sequences is equal to the sum of their limits. A counter example is a specific case where this concept does not hold true.

Why is proving this concept important in mathematics?

Proving the sum of converging sequences is equal to the sum of their limits is important because it helps to establish the validity of mathematical principles and theorems. It also allows us to better understand the behavior of sequences and their limits.

What are some common techniques used to prove this concept?

Some common techniques used to prove the sum of converging sequences are equal to the sum of their limits include the squeeze theorem, the Cauchy criterion, and the limit laws.

Is there a specific condition that needs to be met for this concept to hold true?

Yes, the two sequences must converge to their respective limits in order for this concept to hold true. If one or both of the sequences do not converge, then the sum of the sequences may not be equal to the sum of their limits.

Can you provide an example of a counter example for this concept?

One example of a counter example for this concept is the sequences (1/n) and (-1/n). Both of these sequences converge to 0, but the sum of these sequences is not equal to 0. Therefore, this is a counter example to the concept that the sum of converging sequences is equal to the sum of their limits.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
574
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
492
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
997
  • Calculus and Beyond Homework Help
Replies
2
Views
416
  • Calculus and Beyond Homework Help
Replies
3
Views
905
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
4K
Back
Top