Proving the Validity of My Mathematical Proof

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Homework Help Overview

The discussion revolves around the validity of a mathematical proof concerning convergence and divergence of sequences. Participants are examining the nuances of boundedness, oscillation, and the implications of certain definitions in the context of sequences in mathematical analysis.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants are questioning the assumptions related to convergence and divergence, particularly whether non-diverging sequences can be merely bounded or oscillating. There is also discussion about the redundancy of certain elements in the proof and the implications of contradictions arising from assumptions.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the proof's structure and assumptions. Some guidance has been offered regarding the interpretation of convergence and divergence, but there is no explicit consensus on the proof's validity yet.

Contextual Notes

Participants are navigating the complexities of definitions in mathematical proofs, particularly in relation to sequences, and are considering the implications of boundedness and oscillation on convergence.

Unassuming
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Does my "proof" work?

n\in \mathbb{N}
\).
 
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not diverging may not be equivalent to converging. it may be merely bounded but oscillating. so i would contradict boundedness rather convergence. maybe by showing that the difference between the nth and the n+k is larger than any bound.
 


xaos said:
not diverging may not be equivalent to converging. it may be merely bounded but oscillating.
But such a sequence is certainly diverging / not converging.


About the proof presented in the OP: isn't your N_2 redundant? After all, n+1 > n >= N_1. Other than this minor quibble, your proof is fine.
 


morphism said:
But such a sequence is certainly diverging / not converging.


About the proof presented in the OP: isn't your N_2 redundant? After all, n+1 > n >= N_1. Other than this minor quibble, your proof is fine.

I thought something was wrong with my proof. I was wondering whether N_2 was redundant. I can repeat the definition using N_1 again, or N.

I was wondering about the contradiction though. Does it really contradict? I proved the opposite based off the assumption, therefore I can assume the original is correct?
 


You showed that for any convergent sequence (a_n), the quantity |a_{n+1} - a_n| can be made arbitrarily small (by choosing sufficiently large n). This means that a sequence for which this cannot happen is divergent.
 


ahh, thank you
 


yeah my first guess is usually wrong. you would think that divergence is always unbounded.
 

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