The sequence (aₙ) = (−1)ⁿ does not converge. Which subsequence-based argument best demonstrates this?
AThe sequence is bounded, so by Bolzano-Weierstrass it cannot converge
BThe even-indexed subsequence a₂ₖ converges to 1 and the odd-indexed subsequence a₂ₖ₊₁ converges to −1; since these limits differ, the original sequence cannot converge
CThe sequence has no convergent subsequences, which is sufficient to prove divergence
DThe sequence is not eventually monotone, so it fails the criterion for convergence in ℝ
If a sequence converges to L, every subsequence must converge to L. So if two subsequences converge to different limits, no single L can satisfy the definition — the original sequence cannot converge. The even-indexed terms of (−1)ⁿ are all 1, converging to 1; the odd-indexed terms are all −1, converging to −1. These limits are different, which is a contradiction with any proposed limit L. This is the standard proof technique: find two subsequences with distinct limits to certify divergence.
Question 2 Multiple Choice
Suppose every convergent subsequence of a bounded sequence (aₙ) converges to the same value L. Does (aₙ) necessarily converge to L?
ANo — a sequence can have all subsequences converge to L while the original sequence diverges, as long as some terms wander far from L
BYes — if every convergent subsequence converges to L, then (aₙ) itself converges to L
COnly if the sequence is monotone in addition to being bounded
DOnly if L = 0, since non-zero limits require stronger conditions
If (aₙ) did not converge to L, there would exist an ε > 0 and infinitely many terms with |aₙ − L| ≥ ε. Those infinitely many terms form a subsequence that stays away from L. Since (aₙ) is bounded, Bolzano-Weierstrass guarantees this 'bad' subsequence has a convergent sub-subsequence — but that sub-subsequence converges to some limit other than L (since its terms are all at distance ≥ ε from L). This contradicts the assumption that every convergent subsequence converges to L. Therefore (aₙ) must converge to L.
Question 3 True / False
If a sequence (aₙ) diverges, then no subsequence of (aₙ) can converge.
TTrue
FFalse
Answer: False
This is a common and important misconception. A divergent sequence can have many convergent subsequences — it just cannot have all of them converge to the same limit (which would force convergence of the full sequence). The sequence (−1)ⁿ diverges, yet a₂ₖ → 1 and a₂ₖ₊₁ → −1 are both convergent subsequences. More dramatically, Bolzano-Weierstrass guarantees that every bounded sequence (whether convergent or not) has at least one convergent subsequence. Divergence means the whole sequence fails to settle on a limit, not that it has no structured parts.
Question 4 True / False
If a sequence (aₙ) converges to L, then the subsequence formed by taking only the even-indexed terms (a₂, a₄, a₆, ...) must also converge to L.
TTrue
FFalse
Answer: True
This is a direct application of the theorem that convergence implies every subsequence converges to the same limit. The even-indexed subsequence (a₂ₖ) has indices n₁=2 < n₂=4 < n₃=6 < ..., which is a valid strictly increasing sequence of indices. By definition, for any ε > 0, there exists N such that all n ≥ N satisfy |aₙ − L| < ε. Since nₖ = 2k → ∞, eventually every nₖ ≥ N, so |a₂ₖ − L| < ε for all sufficiently large k. The same argument applies to any subsequence, not just even-indexed ones.
Question 5 Short Answer
Explain why the existence of two subsequences converging to different limits proves that the original sequence diverges.
Think about your answer, then reveal below.
Model answer: The theorem states: if aₙ → L, then every subsequence converges to L. Contrapositive: if some subsequence does NOT converge to L, then aₙ does not converge to L. If two subsequences converge to different values L₁ ≠ L₂, then no matter what value L is proposed as the limit of the original sequence, at least one subsequence (the one converging to the other value) fails to converge to L. Since the original sequence would require all subsequences to agree on L, and none can, the original sequence cannot converge to any limit.
This argument uses the contrapositive of a key theorem and is the standard technique for proving divergence via subsequences. It is more elegant than ε-N arguments for oscillating sequences because it reduces the problem to showing that two parts of the sequence settle near different values — often geometrically obvious. The method also reveals the structure of divergence: the sequence is not 'going to infinity' but is being pulled in multiple directions simultaneously, which is precisely what subsequences with different limits capture.