- #1
Unassuming
- 167
- 0
Does my "proof" work?
n\in \mathbb{N}
\).
n\in \mathbb{N}
\).
Last edited:
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.
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