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What is the logical structure of the proof that the sequence an=(−1)na_n = (-1)^n does not converge?

The proof uses the negation of the formal definition of convergence. It establishes that for every real number LL, there exists an ε>0\varepsilon > 0 (specifically ε=1\varepsilon=1) such that for every natural number NN, there exists an index n>Nn > N where ∣an−L∣≥ε|a_n - L| \ge \varepsilon. This is demonstrated by showing that the odd and even subsequences have distinct limits (−1-1 and 11), and the distance between these limits ensures that no single LL can be close to all terms eventually.

Conditions

  • The sequence is an=(−1)na_n = (-1)^n.
  • The definition of convergence is ∀ε>0,∃N,∀n>N:∣an−L∣<ε\forall \varepsilon > 0, \exists N, \forall n > N: |a_n - L| < \varepsilon.
  • The negation is ∀L,∃ε>0,∀N,∃n>N:∣an−L∣≥ε\forall L, \exists \varepsilon > 0, \forall N, \exists n > N: |a_n - L| \ge \varepsilon.

Reasoning, step by step

  1. State the formal definition of convergence for a sequence to a limit LL.
  2. Negate this definition to formulate the condition for non-convergence.
  3. Identify ε=1\varepsilon = 1 as a witness for the negation.
  4. Use the triangle inequality or subsequence properties to show that for any LL, either the even or odd terms stay at distance ≥1\ge 1 from LL.
  5. Conclude that the negated condition holds for all LL, proving the sequence does not converge.

Example

The card 'Logical Negation Structure' explains: 'Failure to converge to a fixed L differs from having no limit at all: the latter must hold for every real L. Here ε=1ε=1 and the distance 2 between odd and even terms rule out all candidates.' Formula: ∀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.

Common misconceptions

  • Thinking that proving divergence from one specific LL proves the sequence has no limit.
  • Confusing the order of quantifiers in the definition of convergence.
  • Believing that oscillating sequences always diverge without checking if they settle into a pattern.

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