Limit Superior and Limit Inferior — Mathematics Knowledge System

Overview

Not every sequence of real numbers converges, but every bounded sequence has two canonical 'boundary values' that always exist: the limit superior (limsup) and limit inferior (liminf). The limsup is the largest value that infinitely many terms of the sequence get arbitrarily close to (the largest subsequential limit); the liminf is the smallest such value. When these two values coincide, the sequence converges to their common value; when they differ, the sequence oscillates between (at least) two distinct accumulation points and diverges. Limsup and liminf extend to unbounded sequences by allowing the values +∞ or −∞, so they are always defined for any real sequence — unlike the ordinary limit.

Intuition

Imagine standing far out along the sequence, past index N, and asking: what is the highest 'ceiling' the remaining terms aₙ, aₙ₊₁, ... ever bump up against? That ceiling is sup{aₖ : k≥N}. As N grows, you're looking at a shrinking tail of the sequence, so this ceiling can only go down (or stay the same) — it is a decreasing sequence of suprema. The limit of that decreasing sequence of ceilings, as N→∞, is the limsup: the eventual, unavoidable ceiling the tail keeps bumping against no matter how far out you go. Symmetrically, the liminf is the eventual unavoidable floor. A sequence converges exactly when its ceiling and floor squeeze together to the same value.

Formal Definition

Definition

For a sequence (aₙ) of real numbers, define for each N the tail supremum and tail infimum, then take their limits as N→∞ (both limits always exist in the extended reals ℝ ∪ {±∞}, since the tail suprema are decreasing and the tail infima are increasing):

\limsup_{n\to\infty} a_n = \lim_{N\to\infty}\Big(\sup_{n\ge N} a_n\Big) = \inf_{N} \sup_{n \ge N} a_n

The limit of the (non-increasing) sequence of tail suprema

Limit superior

\liminf_{n\to\infty} a_n = \lim_{N\to\infty}\Big(\inf_{n\ge N} a_n\Big) = \sup_{N} \inf_{n \ge N} a_n

The limit of the (non-decreasing) sequence of tail infima

Limit inferior

\liminf_{n\to\infty} a_n \ \le\ \limsup_{n\to\infty} a_n

Ordering (always holds)

\lim_{n\to\infty} a_n = L \iff \liminf_{n\to\infty} a_n = \limsup_{n\to\infty} a_n = L

A sequence converges if and only if its limsup and liminf coincide, in which case both equal the ordinary limit

Convergence criterion

Notation

Notation	Meaning
$\limsup_{n\to\infty} a_n$	The largest subsequential limit of (aₙ); also written \overline{\lim} a_n
$\liminf_{n\to\infty} a_n$	The smallest subsequential limit of (aₙ); also written \underline{\lim} a_n
$\sup_{n \ge N} a_n$	The supremum of the tail of the sequence starting at index N
$\lim_{k\to\infty} a_{n_k}$	The limit of some convergent subsequence of (aₙ); limsup and liminf are the largest and smallest such values

Derivation

Why the tail suprema Mₙ = sup_{k≥N} aₖ form a non-increasing sequence, guaranteeing limsup always exists (possibly as +∞ or −∞).

M_N = \sup_{k \ge N} a_k

Define the tail supremum starting at index N

\{a_k : k \ge N+1\} \subseteq \{a_k : k \ge N\}

The tail starting at N+1 is a subset of the tail starting at N (one fewer term, aₙ, included)

M_{N+1} = \sup_{k\ge N+1} a_k \ \le\ \sup_{k \ge N} a_k = M_N

Taking sup over a smaller set can only decrease (or preserve) the supremum

(M_N) \text{ is non-increasing} \implies \lim_{N\to\infty} M_N \text{ exists in } \mathbb{R}\cup\{-\infty\}

A non-increasing sequence (bounded below by -∞) always has a limit by the Monotone Convergence Theorem (extended to allow -∞)

Proofs

limsup aₙ = liminf aₙ = L implies aₙ → L

$\text{Suppose } \limsup a_n = \liminf a_n = L \in \mathbb{R}.$ (Given hypothesis)
$\text{Let } \varepsilon>0. \text{ Since } \lim_N \sup_{n\ge N} a_n = L, \exists N_1: \sup_{n\ge N_1} a_n < L+\varepsilon.$ (Convergence of the tail-supremum sequence to L)
$\text{Similarly } \exists N_2: \inf_{n\ge N_2} a_n > L - \varepsilon.$ (Convergence of the tail-infimum sequence to L)
$\text{Let } N=\max(N_1,N_2). \text{ For } n\ge N: L-\varepsilon < a_n < L+\varepsilon.$ (For n≥N, aₙ lies between the tail infimum and tail supremum bounds established above)
$\therefore |a_n - L| < \varepsilon \text{ for all } n \ge N, \text{ so } a_n \to L.$ (Exactly the ε-N definition of convergence)

Properties

Both always exist (extended reals)

\limsup a_n, \ \liminf a_n \in \mathbb{R} \cup \{+\infty,-\infty\} \text{ for every real sequence } (a_n)

Condition: Unlike the ordinary limit, which may fail to exist, limsup and liminf are always defined — they equal ±∞ precisely when the sequence is unbounded above/below.

Extreme subsequential limits

\limsup a_n = \max\{\text{subsequential limits of } (a_n)\},\quad \liminf a_n = \min\{\text{subsequential limits of } (a_n)\}

Condition: By the Bolzano–Weierstrass theorem, a bounded sequence has at least one subsequential limit, so this max/min is well-defined.

Squeeze / convergence criterion

a_n \to L \iff \limsup a_n = \liminf a_n = L

Subadditivity

\limsup (a_n+b_n) \le \limsup a_n + \limsup b_n

Condition: Strict inequality can occur, e.g. when aₙ and bₙ oscillate out of phase; equality holds if one of the two sequences converges.

Reversal under negation

\liminf(-a_n) = -\limsup a_n

Applications

The ratio test's more powerful cousin, the root test, uses limsup |aₙ|^(1/n) to determine radius of convergence of power series even when the ordinary limit of the ratio fails to exist — essential in analyzing algorithm generating functions.

Worked Examples

The sequence only ever takes the values +1 (even n) and -1 (odd n), each infinitely often.
$a_n \in \{-1, 1\} \text{ for all } n$
The subsequential limits are exactly +1 and -1 (the two constant subsequences).
$\{\text{subsequential limits}\} = \{-1, 1\}$

Answer: limsup aₙ = 1, liminf aₙ = -1 (they differ, confirming the sequence diverges)

Practice Problems

Difficulty 5/10

Find limsup and liminf of aₙ = (-1)ⁿ/n.

Difficulty 6/10

If limsup aₙ = 3 and liminf aₙ = -1, which statement is correct?

aₙ converges to 3aₙ converges to (3 + (-1))/2 = 1aₙ diverges, oscillating with subsequential limits ranging between -1 and 3aₙ is unbounded

Difficulty 8/10

Prove that liminf(-aₙ) = -limsup(aₙ) for any bounded sequence (aₙ).

Common Mistakes

Common Mistake

Believing limsup aₙ must be a subsequential limit that the sequence 'approaches from above' only, or confusing it with just the supremum of the whole sequence.

limsup is the LARGEST subsequential limit, not the overall supremum. E.g. for aₙ = 1/n for n odd and aₙ=1-1/n for n even, sup aₙ could be close to 1 from a single early term, but limsup is about the eventual, persistent ceiling — here limsup = 1 (from the even subsequence) even if some individual early odd term happens to be larger.

Common Mistake

Assuming limsup and liminf require the sequence to be bounded to be defined.

They are defined for every real sequence, using the extended reals ℝ∪{±∞}. An unbounded-above sequence simply has limsup = +∞, rather than being undefined.

Common Mistake

Thinking a sequence with limsup = liminf must be monotone.

Convergence (limsup=liminf) says nothing about monotonicity — an oscillating sequence like (-1)ⁿ/n has limsup=liminf=0 and converges, without ever being monotone.

Quiz

Flashcards

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Summary

limsup aₙ = lim_{N→∞} sup_{n≥N} aₙ and liminf aₙ = lim_{N→∞} inf_{n≥N} aₙ; both always exist in the extended reals ℝ∪{±∞}, even when the ordinary limit does not.
limsup and liminf are the largest and smallest subsequential limits of the sequence, respectively.
A sequence converges to L if and only if limsup aₙ = liminf aₙ = L.
For aₙ = (-1)ⁿ(1+1/n): the even subsequence decreases to 1 (limsup=1) and the odd subsequence increases to -1 (liminf=-1).
Because limsup/liminf are always defined, they extend tools like the root test to sequences whose ordinary limits fail to exist.

References

WebsiteWikipedia — Limit superior and limit inferior
BookRudin, W. Principles of Mathematical Analysis, 3rd ed. Ch. 3.