Prove: Advanced Calc Proof: ((a(n))+(a(n+1)))/2 Converges to A

  • Thread starter Thread starter bloynoys
  • Start date Start date
  • Tags Tags
    advanced Proof
Click For Summary

Homework Help Overview

The discussion revolves around the convergence of sequences, specifically examining the conjecture that if the average of two consecutive terms of a sequence converges to a limit A, then the sequence itself must also converge to A. Participants are exploring the implications of this conjecture and related statements about boundedness in sequences of real numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to prove the conjecture regarding the convergence of the sequence based on the average of its terms. Some participants question the validity of the conjecture by providing counterexamples, such as the alternating sequence of 1 and -1. Others raise related conjectures about boundedness in sequences and discuss the implications of the phrasing of these conjectures.

Discussion Status

Participants are actively engaging with the conjectures, providing examples and counterexamples to clarify their understanding. There is a recognition of the complexity involved in proving or disproving the conjectures, and some guidance is offered regarding the interpretation of the statements.

Contextual Notes

Participants express frustration with the abstract nature of proof-based classes and the challenges of applying theoretical concepts to specific problems. The discussion highlights the importance of understanding the nuances in the definitions and implications of convergence and boundedness.

bloynoys
Messages
25
Reaction score
0

Homework Statement


If ((a(n))+(a(n+1)))/2 converges to A, then a(n) converges to A. Either prove or disprove this conjecture.



Homework Equations


Normal convergence proof



The Attempt at a Solution


I will prove that ((a(n))+(a(n+1)))/2 converges to A such that for every (epsilon)>0 there exists a positive integer N such that for every n>N abs(a(n)-A)<(epsilon).

Consider (epsilon)> 0 arbitrary.
Because a(n) and a(n+1) will converge to the same number...

That second part is what I am stuck at. I am really good at proving things converge this way with sequences defined but struggle in the abstracts. I struggle with the second and fourth lines of the proof, as the third is just consider n>N arbitrary.
 
Physics news on Phys.org
Consider the sequence 1, -1, 1, -1, ...that is a(n)=1 if n is odd, a(n)=-1 if n is even. What is the limit of (a(n)+a(n+1))/2?ehild
 
Wow, yeah. And I just used that to disprove the question before that. Thank you!

Another related question on proofs:

One is false and one is true (Prove the true, disprove the false)

Conjecture A: If a(n) is a sequence of real numbers then for every positive integer n there exists an M such that abs(a(n)) <= M.

Conjecture B: If a(n) is a sequence of real numbers there exists an M such that every positive integer n abs(a(n)) <= M.

I have been staring at it and I should know it but am struggling to figure it out.
 
bloynoys said:
Conjecture A: If a(n) is a sequence of real numbers then for every positive integer n there exists an M such that abs(a(n)) <= M.

Conjecture B: If a(n) is a sequence of real numbers there exists an M such that every positive integer n abs(a(n)) <= M.
I think you miss something from sentence B. abs(a(n)) <= M has to be true for every positive integer n.

Think what do the sentences mean? Imagine a simple sequence, 0.1, 0.2, 0.3,... for example. Is A true? Choose an n, for example n=1000. Can you find an M so as |an|<=M?

Can you find an M so as |a(n)|<=M for every n?

ehild
 
Thank you! That is what so frustrating about an entirely proof based class at this level. There is so many ways to look at these things and you understand it in class but when you get home to do questions that are slightly different you just sit there, as I know the stuff, but helps to have someone saying consider this... and then connecting the dots.
 
sometimes it helps to know what is really being said.

conjecture A states that every term of a real sequence is a finite real number.

conjecture B states that every real sequence is bounded.

put this way, it should be clear which one is true, and which one is false.

(and, on a more basic level, why the order of "for all" and "there exists" can't simply be reversed).
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K