Questions: Limit Superior and Inferior

Question 1

Consider the sequence aₙ = (−1)ⁿ + 1/n. What is lim sup aₙ?

• A. +∞, because the sequence is unbounded
• B. 0, because 1/n → 0 and that is the only term that changes
• C. 1, because along even-indexed terms aₙ = 1 + 1/n → 1, and this is the largest value approached infinitely often
• D. The limit superior does not exist for oscillating sequences

Answer: (C)

For even n, aₙ = 1 + 1/n, which approaches 1 from above. For odd n, aₙ = −1 + 1/n → −1. The tail supremum sup_{k≥n} aₖ is approximately 1 + 1/n (the even-indexed terms dominate), and this approaches 1 as n → ∞. So lim sup aₙ = 1. The key: lim sup is the largest value the sequence approaches infinitely often — not the largest value ever achieved. The lim inf = −1.

Question 2

A student claims 'lim sup aₙ is just the supremum of the sequence — the largest value aₙ ever takes.' Which sequence disproves this?

• A. aₙ = (−1)ⁿ — here sup = 1 and lim sup = 1, so they agree
• B. aₙ = 1/n — here sup = 1 (the first term) but lim sup = 0, since 1 appears only finitely often and the sequence converges to 0
• C. aₙ = n — here both sup and lim sup equal +∞, so they agree
• D. aₙ = sin(n) — here both sup and lim sup equal 1

Answer: (B)

For aₙ = 1/n, the supremum is a₁ = 1, but lim sup aₙ = 0. The tail supremum sup_{k≥n} aₖ = 1/n → 0. The value 1 appears only once (at n=1) — lim sup ignores finitely-occurring values. It captures only the largest value the sequence approaches infinitely often — the largest limit point of any subsequence. This is fundamentally different from supremum, which is a one-time-maximum.

Question 3

A sequence (aₙ) converges to L if and only if lim sup aₙ = lim inf aₙ = L.

• T. True
• F. False

Answer: True

This is the fundamental characterization that makes lim sup and lim inf strictly more general than ordinary limits. If (aₙ) converges to L, every tail supremum and infimum approaches L, so limsup = liminf = L. Conversely, if limsup = liminf = L, the sequence is squeezed: for large n, all subsequent terms lie in an arbitrarily small interval around L, which is exactly ε-N convergence. Ordinary limits are the special case where the two extremes coincide.

Question 4

For a bounded sequence, lim sup aₙ equals the supremum of most values the sequence takes.

• T. True
• F. False

Answer: False

The lim sup is the largest LIMIT POINT (subsequential limit) — the largest value the sequence approaches infinitely often. The overall supremum may be achieved at just finitely many terms and then never approached again. For example, aₙ = 1/n has sup = 1 (attained at n=1) but lim sup = 0 (the only limit point). The lim sup effectively ignores transient large values and tracks only persistent accumulation points.

Question 5

Why do the limit superior and inferior always exist for bounded sequences, even when the ordinary limit does not?

Model answer: Define Mₙ = sup_{k≥n} aₖ (the tail supremum). As n increases, we're taking sup over a smaller set, so Mₙ is non-increasing. Since the sequence is bounded below, the non-increasing sequence (Mₙ) is also bounded below — by the monotone convergence theorem, it converges. This limit is the lim sup. Symmetrically, the tail infima mₙ = inf_{k≥n} aₖ form a non-decreasing bounded sequence that converges; this is the lim inf. The ordinary limit fails when these two limits are different (the sequence oscillates between multiple accumulation points), but each individually always converges.

This is why lim sup and lim inf are defined via tail operations and limits, not directly as 'largest value approached.' The construction via monotone sequences guarantees existence. The ordinary limit is just the special case where Mₙ and mₙ converge to the same value — a coincidence that fails for oscillating sequences.