For the sequence aₙ = (−1)ⁿ + 1/n, what are the limsup and liminf?
Alimsup = 1, liminf = −1
Blimsup = 1, liminf = −1, and therefore the sequence converges
Climsup and liminf do not exist because the sequence does not converge
Dlimsup = 2, liminf = 0
The sequence oscillates between values near +1 (odd terms: −1 + 1/n → −1) wait — even terms give +1 + 1/n → 1, odd terms give −1 + 1/n → −1. So the sequence approaches 1 and −1 infinitely often. limsup = 1 (the largest accumulation point) and liminf = −1 (the smallest). Since limsup ≠ liminf, the sequence diverges. Option C reflects the common misconception that limsup/liminf require convergence — they are defined precisely for sequences that do *not* converge. Option B wrongly concludes convergence from the values.
Question 2 Multiple Choice
A bounded sequence has limsup aₙ = liminf aₙ = 5. What can you conclude?
AThe sequence is eventually constant, equal to 5
BThe sequence converges to 5
CThe sequence has 5 as its supremum but may not converge
DThe sequence converges to 5 only if it is monotone
The fundamental theorem connecting limsup, liminf, and convergence states: a bounded sequence converges if and only if its limsup equals its liminf, in which case both equal the ordinary limit. If limsup = liminf = 5, the sequence converges to 5 — by definition, the tail suprema and infima both squeeze to 5, trapping all eventual terms arbitrarily close to 5. The sequence need not be eventually constant (e.g., aₙ = 5 + sin(n)/n satisfies this). Monotonicity is irrelevant; the squeeze is the whole argument.
Question 3 True / False
For any bounded sequence, limsup aₙ ≥ liminf aₙ.
TTrue
FFalse
Answer: True
This follows directly from the definitions. For every n, sup{aₖ : k ≥ n} ≥ inf{aₖ : k ≥ n} (the supremum of a set is always ≥ its infimum). Taking limits preserves this inequality: the decreasing sequence Sₙ converges to limsup, the increasing sequence Iₙ converges to liminf, and Sₙ ≥ Iₙ for every n implies limsup ≥ liminf. Equality is the special case of convergence; strict inequality signals oscillation.
Question 4 True / False
If a bounded sequence does not converge, its limsup and liminf do not exist.
TTrue
FFalse
Answer: False
This is the key misconception. Limsup and liminf are *always* defined for any bounded sequence — that is their entire purpose. The sequences Sₙ = sup{aₖ : k ≥ n} and Iₙ = inf{aₖ : k ≥ n} are monotone (decreasing and increasing respectively) and bounded, so the Monotone Convergence Theorem guarantees their limits exist. The ordinary limit fails for an oscillating sequence precisely because limsup ≠ liminf; but both individual values exist. Limsup and liminf are tools for sequences that fail to converge.
Question 5 Short Answer
The root test for series uses limsup |aₙ|^(1/n) rather than lim |aₙ|^(1/n). Why is the limsup version more useful?
Think about your answer, then reveal below.
Model answer: The ordinary limit lim |aₙ|^(1/n) may not exist for all sequences, but limsup always exists for bounded sequences. If the root test used lim, it would be inapplicable whenever the sequence oscillates or has irregular behavior. Using limsup makes the test universally applicable: whenever limsup |aₙ|^(1/n) < 1, the series converges absolutely, regardless of whether the ordinary limit exists. The limsup captures the 'worst-case eventual growth rate' of the terms, which is exactly what determines convergence.
This is the practical payoff of limsup and liminf: they are precision tools for exactly the irregular, non-convergent sequences where blunter tools like ordinary limits fail. The root test is most needed precisely when |aₙ|^(1/n) oscillates — and limsup is defined even then.