Why is rejecting just the candidates $L=1$ and $L=-1$ not enough to prove divergence for $a_n = (-1)^n$?

Conditions

• The sequence is defined as $a_n = (-1)^n$.
• Proof requires ruling out all $L \in \mathbb{R}$.

Reasoning, step by step

1. Assume a hypothetical limit $L$ exists.
2. Apply the triangle inequality: $2 = |1 - (-1)| \le |1 - L| + |-1 - L|$.
3. Deduce that at least one of the distances $|1-L|$ or $|-1-L|$ must be $\ge 1$.
4. Choose $\varepsilon = 1$. The subsequence corresponding to the farther point will always have terms outside $(L-\varepsilon, L+\varepsilon)$.
5. Conclude that no single $L$ can satisfy the condition for all sufficiently large $n$.

Example

If we pick $L=0$, both 1 and -1 are at distance 1. With $\varepsilon=1$, neither is strictly inside the band $( -1, 1 )$ if boundaries are exclusive, or they sit on the edge depending on strictness, but typically we use $\varepsilon < 1$ or rely on the fact that for any $L$, one cluster is far away.

Common misconceptions

• Believing that checking the visible oscillation points (1 and -1) covers all possibilities.
• Confusing 'limit does not exist' with 'limit is infinite'.

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• How does the formal negation of the limit definition apply here?
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