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What is the formal logical statement representing the divergence of an=(−1)na_n=(-1)^n based on the negated limit definition?

The divergence is proven by asserting: For every real number LL, there exists an ε>0\varepsilon > 0 such that for every integer NN, there is an index n>Nn > N where ∣an−L∣≥ε|a_n - L| \ge \varepsilon.

Conditions

  • Standard definition of sequence convergence is assumed.
  • an=(−1)na_n = (-1)^n.

Reasoning, step by step

  1. Start with the convergence definition: ∀ε>0,∃N,∀n>N,∣an−L∣<ε\forall \varepsilon > 0, \exists N, \forall n > N, |a_n - L| < \varepsilon.
  2. Negate the quantifiers and inequality to define non-convergence.
  3. Swap ∀\forall and ∃\exists appropriately.
  4. Change strict inequality << to ≥\ge.
  5. Result: ∀L∈R,∃ε>0,∀N,∃n>N:∣an−L∣≥ε\forall L \in \mathbb{R}, \exists \varepsilon > 0, \forall N, \exists n > N : |a_n - L| \ge \varepsilon.

Example

The video card displays this exact formula: '∀L∈R\forall L\in\mathbb R,\ ∃ε>0\exists\varepsilon>0,\ ∀N\forall N,\ ∃n>N\exists n>N:\ |an−La_n-L|≥ε\ge\varepsilon'. This means no matter what limit you guess, I can find a bad epsilon that trips you up infinitely often.

Common misconceptions

  • Thinking divergence means ∣an−L∣→∞|a_n - L| \to \infty.
  • Forgetting that the negation applies to *all* possible limits LL, not just one.

Watch the explanation

1:18 – 1:34Watch this moment ↗

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.