Simpler proof that sequence x_n = n does not converge?

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Discussion Overview

The discussion revolves around proving that the sequence \{n\}_{n\in \mathbb{N}} does not converge, focusing on the definition of convergence and exploring different proof strategies. Participants consider the complexity of the proof and seek simpler or more elegant approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a proof by contradiction, using the Archimedean Principle to argue that if the sequence converges, it leads to a contradiction.
  • Another participant suggests that a simpler proof could be based on the property that convergent sequences are bounded, referencing the triangle inequality.
  • A different viewpoint emphasizes that while the original proof may seem long, it effectively translates a heuristic argument into a rigorous mathematical form.
  • Participants discuss the nature of proofs, noting that what seems obvious may still require detailed justification in formal mathematics.

Areas of Agreement / Disagreement

Participants express differing opinions on the complexity of the original proof, with some finding it straightforward while others perceive it as unnecessarily complicated. No consensus is reached on a simpler proof method.

Contextual Notes

Participants acknowledge the potential for missing details in heuristic arguments and the challenges of rigorously stating such arguments in natural language.

Who May Find This Useful

Readers interested in mathematical proofs, particularly in the context of sequences and convergence, may find this discussion relevant.

brian44
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Hi, I am trying to prove that
[itex] \{n\}_{n\in \mathbb{N}}[/itex]
does not converge (based on definition of convergence).

I can prove this by contradiction saying assume it converges, fix [itex]\epsilon[/itex] , then [itex]x_n < \epsilon + a[/itex] (for [itex]n \ge N[/itex] where N is fixed) (by fundamental theorem of ineq.) but by Archimedean Principle, I can find a natural number that surpasses this bound, i.e. [itex]\exists m , m x_n > \epsilon + a[/itex] which is an element of the sequence [itex]x_n[/itex] which means for some M, [itex]n \ge M \rightarrow x_n > \epsilon + a[/itex] which is a contradiction.

However this seems like a long complicated proof for a very simple and obvious fact, I was wondering if there is not some easier, more elegant way to prove this that I am missing?

Thanks for your help.
-Brian
 
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How about saying that a convergent sequence is bounded? It is a simple exercise using the triangle inequality.
 
brian44 said:
However this seems like a long complicated proof for a very simple and obvious fact
When you have limited tools at your disposal, that tends to make proofs long and complicated. (and obvious facts tend to be either very easy to prove or very tricky to prove)

However, I'm not really convinced that your proof is all that long and complicated. It's only what? Three lines long? And the proof boils down to nothing more than:
If it converges, it would eventually have to stop growing. However, it keeps growing forever. Therefore, it doesn't converge​
does it not?

I assert the proof you gave is a direct translation of the above heuristic argument into a rigorous mathematical argument.

Actually, if you notice, the heuristic argument I gave is full of little missing details that would be very cumbersome to state in natural language. (e.g. a sequence can grow forever but still converge. So I really mean something about how fast it grows) But in this case, those details are very easy to state in the mathematical language!
 
Thanks for the feedback.

And I guess it isn't that long, it was just in my head as I was writing it out going through the process I was thinking "I can't believe I have to write all this for something that is so obvious."
 

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