Unassuming
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Does my "proof" work?
n\in \mathbb{N}
\).
n\in \mathbb{N}
\).
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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.
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.
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.
But such a sequence is certainly diverging / not converging.xaos said:not diverging may not be equivalent to converging. it may be merely bounded but oscillating.
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.