Limit Superior and Limit Inferior

Consider a sequence that refuses to converge, such as an=(−1)n(1+1n)a_n = (-1)^n\left(1 + \tfrac{1}{n}\right). It oscillates forever. Yet its long-run behavior is not chaotic: the odd terms creep down toward −1-1. The even terms creep down toward 11. We would like a way to capture this "eventual boundary behavior" with real numbers. This should work even when the ordinary limit does not exist. The limit superior and limit inferior do exactly this. They extract the largest and smallest subsequential limits of any bounded sequence. They do this for every sequence at all, once we allow the values ±∞\pm\infty. This matters because convergence is a fragile, all-or-nothing notion. By contrast, lim sup⁡\limsup and lim inf⁡\liminf are robust: they exist for every sequence of real numbers. They convert questions about convergence into questions about inequalities between two always-well-defined quantities. They also give a clean criterion — lim sup⁡an=lim inf⁡an\limsup a_n = \liminf a_n — for a sequence to converge. This idea rests on the Bolzano-Weierstrass Theorem and on the theory of Subsequences and Their Limits. It will resurface almost immediately in the Root Test. It also appears in computing the Radius of Convergence of a power series. There, the limit of a ratio or root frequently fails to exist, but the limsup always does. The big picture is this: a bounded sequence has a nonempty, bounded set of subsequential limits. That set has a maximum and a minimum. The limit superior is the maximum, the limit inferior is the minimum. Equivalently, and more usefully for computation, lim sup⁡an\limsup a_n is the limit of the shrinking tail-suprema sup⁡k≥nak\sup_{k\ge n} a_k. Similarly, lim inf⁡an\liminf a_n is the limit of the growing tail-infima inf⁡k≥nak\inf_{k \ge n} a_k. These two descriptions — subsequential-limit and tail-monotone-sequence — are proven equivalent below. Each is useful in different arguments. This article gives the rigorous definition of lim sup⁡\limsup and lim inf⁡\liminf. It proves their basic algebraic and order properties. It proves the fundamental convergence criterion, and works through nontrivial numerical examples. It also shows, with an explicit counterexample, why subadditivity of lim sup⁡\limsup can fail to be equality. Finally, it prepares the ground for later criteria in series convergence.

Key takeaways
  • For any real sequence (an)(a_n), define bn=sup⁡k≥nakb_n = \sup_{k\ge n} a_k and cn=inf⁡k≥nakc_n = \inf_{k \ge n} a_k; then (bn)(b_n) decreases and (cn)(c_n) increases, so both have limits in [−∞,∞][-\infty,\infty].
  • lim sup⁡an:=lim⁡nbn\limsup a_n := \lim_n b_n and lim inf⁡an:=lim⁡ncn\liminf a_n := \lim_n c_n; these exist (possibly ±∞\pm\infty) for every sequence, unlike lim⁡an\lim a_n.
  • Equivalently, lim sup⁡an\limsup a_n and lim inf⁡an\liminf a_n are the largest and smallest subsequential limits of (an)(a_n).
  • Always lim inf⁡an≤lim sup⁡an\liminf a_n \le \limsup a_n, with equality iff lim⁡an\lim a_n exists (in the extended sense), and then all three are equal.
  • lim sup⁡\limsup is subadditive (lim sup⁡(an+bn)≤lim sup⁡an+lim sup⁡bn\limsup(a_n+b_n) \le \limsup a_n + \limsup b_n) and lim inf⁡\liminf is superadditive, with strict inequality possible.
  • Key idea: lim sup⁡\limsup and lim inf⁡\liminf tame divergence by tracking the eventual upper and lower envelopes of a sequence, turning "does the limit exist?" into "do two always-defined numbers agree?"

Where this lives today

The limsup/liminf formalism is indispensable wherever an ordinary limit is too fragile to rely on. In the theory of infinite series, the <span class="wiki-link-missing">Root Test</span> states that ∑an\sum a_n converges absolutely if lim sup⁡n∣an∣1/n<1\limsup_n |a_n|^{1/n} < 1 and diverges if this limsup exceeds 11 — using limsup rather than lim because ∣an∣1/n|a_n|^{1/n} frequently oscillates (e.g., for series with alternating block structure) even when the series itself converges nicely. The Cauchy–Hadamard formula for the <span class="wiki-link-missing">Radius of Convergence</span> of a power series ∑cnzn\sum c_n z^n is R=1/lim sup⁡n∣cn∣1/nR = 1/\limsup_n |c_n|^{1/n}, valid unconditionally, whereas the ratio-based version requires the ordinary limit of ∣cn+1/cn∣|c_{n+1}/c_n| to exist. In probability theory, the Borel–Cantelli lemmas and the theory of almost-sure convergence use lim sup⁡\limsup of events (in the set-theoretic sense, lim sup⁡En=⋂n⋃k≥nEk\limsup E_n = \bigcap_n \bigcup_{k\ge n} E_k, the event that infinitely many EnE_n occur) as the direct analogue of the real-number construction. In dynamical systems and ergodic theory, Lyapunov exponents are defined via lim sup⁡\limsup of 1nlog⁡∥Dfn(x)∥\frac{1}{n}\log\|Df^n(x)\| precisely because the ordinary limit is not known to exist a priori. In optimization and control theory, worst-case long-run performance criteria are phrased using lim sup⁡\limsup (e.g. lim sup⁡\limsup average cost), because guarantees must hold no matter how a system's transient oscillations behave. In real analysis itself, lim sup⁡\limsup and lim inf⁡\liminf underlie the definition of upper and lower Riemann/Darboux sums' limiting behavior and appear in the statement of Fatou's lemma in measure theory, ∫lim inf⁡fn dμ≤lim inf⁡∫fn dμ\int \liminf f_n \, d\mu \le \liminf \int f_n\, d\mu, a cornerstone of the Lebesgue integration framework.

Ask the tutor: Why does the Root Test need lim sup⁡\limsup instead of an ordinary limit, and can you show a series where the limit of ∣an∣1/n|a_n|^{1/n} fails to exist but the Root Test still decides convergence?

DefinitionLimit superior and limit inferior

Let (an)n=1∞(a_n)_{n=1}^\infty be a sequence of real numbers. For each n∈Nn \in \mathbb{N} define the tail supremum and tail infimum

bn=sup⁡{ak:k≥n}∈(−∞,+∞],cn=inf⁡{ak:k≥n}∈[−∞,+∞),b_n = \sup\{a_k : k \ge n\} \in (-\infty, +\infty], \qquad c_n = \inf\{a_k : k \ge n\} \in [-\infty, +\infty),

where bn=+∞b_n = +\infty if {ak:k≥n}\{a_k : k\ge n\} is unbounded above, and cn=−∞c_n = -\infty if it is unbounded below. Since the set {ak:k≥n+1}⊆{ak:k≥n}\{a_k : k \ge n+1\} \subseteq \{a_k : k \ge n\}, we have bn+1≤bnb_{n+1} \le b_n and cn+1≥cnc_{n+1} \ge c_n; that is, (bn)(b_n) is nonincreasing and (cn)(c_n) is nondecreasing. A monotone sequence in the extended reals [−∞,∞][-\infty,\infty] always has a limit there (proved below), so we define

lim sup⁡n→∞an:=lim⁡n→∞bn=inf⁡n≥1sup⁡k≥nak,\limsup_{n\to\infty} a_n := \lim_{n\to\infty} b_n = \inf_{n\ge 1}\sup_{k\ge n} a_k,
lim inf⁡n→∞an:=lim⁡n→∞cn=sup⁡n≥1inf⁡k≥nak.\liminf_{n\to\infty} a_n := \lim_{n\to\infty} c_n = \sup_{n\ge 1}\inf_{k\ge n} a_k.

Both quantities exist in the extended real line [−∞,+∞][-\infty,+\infty] for every sequence (an)(a_n).

Intuition

Picture a sequence plotted as points along the horizontal axis, indexed by time going up a vertical axis: at each height nn, mark the point ana_n. The tail supremum bnb_n is the smallest ceiling that bounds all the points at height ≥n\ge n; as nn grows, we are looking further into the future, so the tightest ceiling can only come down (or stay the same). Symmetrically cnc_n is the highest floor at each stage, and it can only rise. The sequence (bn)(b_n) slides down to a limiting ceiling, the lim sup⁡\limsup; the sequence (cn)(c_n) rises to a limiting floor, the lim inf⁡\liminf. Any oscillation of the original sequence gets squeezed between these two converging envelopes. A second mental picture: the set SS of all subsequential limits of a bounded sequence is a nonempty closed bounded set (a consequence of Bolzano-Weierstrass Theorem and a diagonal/closure argument), so it has a largest and smallest element. The claim, proved as our main theorem, is that these are exactly lim sup⁡an\limsup a_n and lim inf⁡an\liminf a_n. For the oscillating sequence an=(−1)n(1+1/n)a_n = (-1)^n(1+1/n), the subsequential limits are exactly {−1,1}\{-1, 1\}, so lim sup⁡an=1\limsup a_n = 1 and lim inf⁡an=−1\liminf a_n = -1.

Figure
Tail sup/inf envelopes squeezing an=(−1)n(1+1/n)a_n=(-1)^n(1+1/n)
n a_n 1 = limsup -1 = liminf a_n points b_n = sup tail ↓ 1 c_n = inf tail ↑ -1

The central theorem

TheoremExistence, subsequential characterization, and convergence criterion

Let (an)(a_n) be a sequence of real numbers, and let bn=sup⁡k≥nakb_n = \sup_{k \ge n} a_k, cn=inf⁡k≥nakc_n = \inf_{k\ge n} a_k as above. Let

L∗=lim sup⁡n→∞an=lim⁡nbn,L∗=lim inf⁡n→∞an=lim⁡ncn,L^* = \limsup_{n\to\infty} a_n = \lim_n b_n, \qquad L_* = \liminf_{n \to \infty} a_n = \lim_n c_n,

taken in the extended reals [−∞,+∞][-\infty, +\infty]. Let S⊆[−∞,+∞]S \subseteq [-\infty,+\infty] denote the set of all subsequential limits of (an)(a_n) (extended-real limits of convergent subsequences, allowing subsequences that diverge to ±∞\pm\infty). Then: 1. (Existence.) L∗L^* and L∗L_* exist in [−∞,+∞][-\infty,+\infty] for every sequence (an)(a_n). 2. (Order.) L∗≤L∗L_* \le L^* always. 3. (Subsequential characterization.) SS is nonempty, L∗=max⁡SL^* = \max S, and L∗=min⁡SL_* = \min S. 4. (Convergence criterion.) The extended-real limit lim⁡nan\lim_n a_n exists (i.e. (an)(a_n) converges to a finite number, or diverges to +∞+\infty, or diverges to −∞-\infty) if and only if L∗=L∗L_* = L^*, in which case lim⁡nan=L∗=L∗\lim_n a_n = L_* = L^*.

Proof

We prove each part in turn using only monotone-sequence facts and the definition of subsequential limit, together with the Bolzano-Weierstrass Theorem.

Part 1: Existence. The sequence (bn)(b_n) is nonincreasing in [−∞,+∞][-\infty, +\infty] (shown in the definition, since tails shrink as nn grows). A nonincreasing sequence in the extended reals always converges in [−∞,+∞][-\infty,+\infty]: if (bn)(b_n) is bounded below by some real mm, then {bn}\{b_n\} is a nonincreasing sequence of reals bounded below, hence converges to β=inf⁡nbn∈R\beta = \inf_n b_n \in \mathbb{R} by the Monotone Convergence Theorem (an ordinary bounded monotone real sequence converges to its infimum, since for any ε>0\varepsilon>0 there is NN with bN<β+εb_N < \beta+\varepsilon, and monotonicity gives β≤bn<β+ε\beta \le b_n < \beta + \varepsilon for all n≥Nn \ge N). If (bn)(b_n) is not bounded below, then for every MM there is NN with bN<−Mb_N < -M, and monotonicity gives bn<−Mb_n < -M for all n≥Nn \ge N, i.e. bn→−∞b_n \to -\infty. If some bn=+∞b_n = +\infty, monotonicity is vacuous at that stage but the same trichotomy (converges to a finite limit, or →−∞\to -\infty) applies to the tail once finite terms appear, or the sequence is eventually +∞+\infty throughout if bn=+∞b_n = +\infty for all nn, in which case L∗=+∞L^* = +\infty by definition. In every case lim⁡nbn\lim_n b_n exists in [−∞,+∞][-\infty, +\infty], so L∗L^* is well defined. The argument for (cn)(c_n) nondecreasing is symmetric (replace sup by inf, bounded below by bounded above), giving existence of L∗L_*.

Part 2: Order. For every nn, cn=inf⁡k≥nak≤sup⁡k≥nak=bnc_n = \inf_{k\ge n} a_k \le \sup_{k \ge n} a_k = b_n, because the infimum of a nonempty set is at most its supremum (indeed cn≤an≤bnc_n \le a_n \le b_n directly). Passing to the limit in nn preserves the non-strict inequality cn≤bnc_n \le b_n (order is preserved by limits, including extended-real limits: if xn≤ynx_n \le y_n for all nn and xn→xx_n \to x, yn→yy_n \to y in [−∞,∞][-\infty,\infty], then x≤yx \le y, by the same ε\varepsilon-argument as for real sequences, extended trivially to infinite limits). Hence L∗≤L∗L_* \le L^*.

Part 3: Subsequential characterization. We show S≠∅S \neq \emptyset and L∗=max⁡SL^* = \max S; the argument for L∗=min⁡SL_* = \min S is symmetric. Case A: L∗=+∞L^* = +\infty. Then bn=sup⁡k≥nak=+∞b_n = \sup_{k\ge n} a_k = +\infty for every nn (since bnb_n is nonincreasing and its limit is +∞+\infty, we cannot have any finite bnb_n, as that would force bm≤bn<∞b_m \le b_n <\infty for m≥nm \ge n, contradicting bm→∞b_m \to \infty). So for every nn and every M>0M > 0 there exists k≥nk \ge n with ak>Ma_k > M. This lets us construct, inductively, indices n1<n2<n3<⋯n_1 < n_2 < n_3 < \cdots with anj>ja_{n_j} > j: having chosen nj−1n_{j-1}, apply the unboundedness of {ak:k≥nj−1+1}\{a_k : k \ge n_{j-1}+1\} with M=jM = j to find nj>nj−1n_j > n_{j-1} with anj>ja_{n_j} > j. Then anj→+∞a_{n_j} \to +\infty, so +∞∈S+\infty \in S, and clearly +∞=max⁡S+\infty = \max S since no extended real exceeds +∞+\infty. Thus L∗=+∞=max⁡SL^* = +\infty = \max S. Case B: L∗=−∞L^* = -\infty. Then bn→−∞b_n \to -\infty, so in particular b1=sup⁡kak<∞b_1 = \sup_k a_k < \infty eventually forces ak→−∞a_k \to -\infty: precisely, since bn→−∞b_n \to -\infty and cn≤bnc_n \le b_n with cnc_n nondecreasing, also bn≥an≥cnb_n \ge a_n \ge c_n; but more directly, for every MM there is NN with bN<−Mb_N < -M, meaning sup⁡k≥Nak<−M\sup_{k\ge N} a_k < -M, so ak<−Ma_k < -M for all k≥Nk \ge N. This says exactly an→−∞a_n \to -\infty as an ordinary (extended) sequence limit. Hence every subsequence also tends to −∞-\infty, so S={−∞}S = \{-\infty\}, nonempty, and L∗=−∞=max⁡SL^* = -\infty = \max S. Case C: L∗L^* is finite. First we show L∗∈SL^* \in S. Fix j∈Nj \in \mathbb{N}. Since bn→L∗b_n \to L^*, choose NjN_j with bNj<L∗+1/jb_{N_j} < L^* + 1/j; also since L∗L^* is the limit of the nonincreasing sequence bnb_n, bn≥L∗b_n \ge L^* for all nn (a nonincreasing sequence's limit is its infimum, hence a lower bound for all terms... wait, infimum is a lower bound, and bnb_n decreases to it, so indeed bn≥L∗b_n \ge L^* for all nn). So L∗≤bNj=sup⁡k≥Njak<L∗+1/jL^* \le b_{N_j} = \sup_{k \ge N_j} a_k < L^* + 1/j. By definition of supremum, there exists kj≥Njk_j \ge N_j with akj>bNj−1/j>L∗−1/ja_{k_j} > b_{N_j} - 1/j > L^* - 1/j (using bNj>L∗b_{N_j} > L^*, wait we need bNj−1/jb_{N_j} - 1/j; since bNj<L∗+1/jb_{N_j} < L^*+1/j this gives akj<L∗+1/ja_{k_j} < L^* + 1/j automatically as akj≤bNja_{k_j}\le b_{N_j}). So we obtain

L∗−1j<akj≤bNj<L∗+1j,L^* - \tfrac1j < a_{k_j} \le b_{N_j} < L^* + \tfrac1j,

i.e. ∣akj−L∗∣<1/j|a_{k_j} - L^*| < 1/j. Choosing NjN_j strictly increasing in jj (possible since we may always increase NjN_j past any prior index, as bn→L∗b_n \to L^* for arbitrarily large nn too) and correspondingly kj≥Njk_j \ge N_j strictly increasing, we get a subsequence akj→L∗a_{k_j} \to L^*. Hence L∗∈SL^* \in S. Next we show L∗L^* is an upper bound for SS. Let ℓ∈S\ell \in S be an extended-real subsequential limit, achieved along anj→ℓa_{n_j} \to \ell. For every nn, all but finitely many njn_j satisfy nj≥nn_j \ge n, hence anj≤bn=sup⁡k≥naka_{n_j} \le b_n = \sup_{k\ge n} a_k for all large jj. Taking j→∞j \to \infty, ℓ≤bn\ell \le b_n (limits preserve non-strict inequalities). This holds for every nn, so ℓ≤inf⁡nbn=L∗\ell \le \inf_n b_n = L^*. Thus every element of SS is ≤L∗\le L^*, and combined with L∗∈SL^* \in S, we conclude L∗=max⁡SL^* = \max S. By the symmetric argument (infima, nondecreasing cnc_n, subsequences approaching from below), L∗=min⁡SL_* = \min S, and in particular S≠∅S \neq \emptyset since it contains L∗L^* (or, in the finite case, one shows directly S≠∅S\ne\emptyset; note also boundedness: if (an)(a_n) is a bounded sequence, S≠∅S \ne \emptyset also follows independently from the Bolzano-Weierstrass Theorem, which guarantees a convergent subsequence — this is consistent with, and gives an alternative route to, the case-C argument above).

Part 4: Convergence criterion. (⇒\Rightarrow) Suppose lim⁡nan=L\lim_n a_n = L exists in the extended-real sense. Then every subsequence also converges to LL (a standard fact: if an→La_n \to L and n1<n2<⋯n_1 < n_2 < \cdots, then for any neighborhood UU of LL, all but finitely many an∈Ua_n \in U, hence all but finitely many anj∈Ua_{n_j} \in U). So S={L}S = \{L\}, and by Part 3, L∗=max⁡S=L=min⁡S=L∗L^* = \max S = L = \min S = L_*. (⇐\Leftarrow) Suppose L∗=L∗=:LL^* = L_* =: L. Recall cn≤an≤bnc_n \le a_n \le b_n for all nn (since cnc_n and bnb_n are the inf and sup of a set containing ana_n). Since cn→Lc_n \to L and bn→Lb_n \to L, the extended-real Squeeze Theorem gives an→La_n \to L. (Proof of the squeeze fact used: if LL is finite, for ε>0\varepsilon > 0 choose NN with L−ε<cnL - \varepsilon < c_n and bn<L+εb_n < L+\varepsilon for n≥Nn \ge N; then L−ε<cn≤an≤bn<L+εL - \varepsilon < c_n \le a_n \le b_n < L + \varepsilon, so ∣an−L∣<ε|a_n - L| < \varepsilon. If L=+∞L = +\infty, for any MM choose NN with cn>Mc_n > M for n≥Nn\ge N, hence an≥cn>Ma_n \ge c_n > M, so an→+∞a_n \to +\infty. The case L=−∞L=-\infty is symmetric using bnb_n.) Hence lim⁡nan=L=L∗=L∗\lim_n a_n = L = L^* = L_*. ■\blacksquare

∎
CorollaryBounded case restated

If (an)(a_n) is a bounded sequence, then L∗,L∗∈RL^*, L_* \in \mathbb{R} are finite, SS is a nonempty compact subset of R\mathbb{R} (bounded, and closed because it equals the set of limit points of the closed bounded range, whose closure argument mirrors the Bolzano–Weierstrass proof), and (an)(a_n) converges if and only if SS is a single point.

Worked examples

Example 1 — a genuinely mixed, non-symmetric sequence. Let

an={3−1n,n≡1(mod3),−2+1n,n≡2(mod3),(−1)n⋅n,n≡0(mod3).a_n = \begin{cases} 3 - \dfrac{1}{n}, & n \equiv 1 \pmod 3,\\[4pt] -2 + \dfrac{1}{n}, & n \equiv 2 \pmod 3,\\[4pt] (-1)^{n} \cdot n, & n \equiv 0 \pmod 3. \end{cases}

We compute L∗=lim sup⁡anL^* = \limsup a_n and L∗=lim inf⁡anL_* = \liminf a_n using the subsequential characterization (Part 3). Subsequence 1: indices n=1,4,7,10,⋯≡1(mod3)n = 1, 4, 7, 10, \dots \equiv 1 \pmod 3. Here an=3−1/n→3a_n = 3 - 1/n \to 3 as n→∞n \to \infty along this subsequence. Subsequence 2: indices n=2,5,8,11,⋯≡2(mod3)n = 2, 5, 8, 11, \dots \equiv 2 \pmod 3. Here an=−2+1/n→−2a_n = -2 + 1/n \to -2. Subsequence 3: indices n=3,6,9,12,⋯≡0(mod3)n = 3, 6, 9, 12, \dots \equiv 0 \pmod 3. Here an=(−1)nna_n = (-1)^n n. Since these nn are multiples of 33, both even and odd multiples occur (n=6n=6 even, n=9n=9 odd, etc.), so this sub-subsequence itself splits: along even multiples of 33 (i.e. n=6,12,18,…n = 6, 12, 18,\dots), an=n→+∞a_n = n \to +\infty; along odd multiples of 33 (i.e. n=3,9,15,…n = 3, 9, 15, \dots), an=−n→−∞a_n = -n \to -\infty. So the full set of subsequential limits is

S={3,−2,+∞,−∞}.S = \{3, -2, +\infty, -\infty\}.

No other subsequential limits exist because every index nn falls into exactly one of these residue classes, and within each class the values are eventually monotone approaching the stated limit (or the class further splits by parity into two monotone pieces, as shown), exhausting all possible convergent (or properly divergent) subsequences: any subsequence has infinitely many terms in at least one of these four "monotone building blocks," and any subsequential limit must be a subsequential limit of one of these blocks (a subsequence of a sequence built from finitely many interleaved pieces has a further subsequence lying entirely in one piece, by the pigeonhole principle, since infinitely many indices must fall into at least one of finitely many classes). By Part 3 of the theorem,

L∗=max⁡S=+∞,L∗=min⁡S=−∞.L^* = \max S = +\infty, \qquad L_* = \min S = -\infty.

We can double check this against the tail-supremum definition directly: for any nn, the tail {ak:k≥n}\{a_k : k \ge n\} contains terms a6m=6ma_{6m}= 6m for 6m≥n6m \ge n, which are unbounded above, so bn=sup⁡k≥nak=+∞b_n = \sup_{k\ge n} a_k = +\infty for every nn, confirming L∗=lim⁡nbn=+∞L^* = \lim_n b_n = +\infty. Symmetrically cn=−∞c_n = -\infty for every nn, confirming L∗=−∞L_* = -\infty. Since L∗≠L∗L_* \ne L^*, Part 4 tells us lim⁡nan\lim_n a_n does not exist even in the extended sense (it is not that an→±∞a_n \to \pm\infty; the sequence is genuinely oscillating without any single extended-real limit).

Example 2 — a sequence built to test subadditivity, with asymmetric data. Let

an=(−1)n+(−1)n+1n,bn=(−1)n+1+2n.a_n = (-1)^n + \frac{(-1)^{n+1}}{n}, \qquad b_n = (-1)^{n+1} + \frac{2}{n}.

First compute lim sup⁡an\limsup a_n and lim inf⁡an\liminf a_n. For even nn: an=1−1n→1a_n = 1 - \frac{1}{n} \to 1. For odd nn: an=−1+1n→−1a_n = -1 + \frac{1}{n} \to -1. So Sa={1,−1}S_a = \{1, -1\}, giving

lim sup⁡an=1,lim inf⁡an=−1.\limsup a_n = 1, \qquad \liminf a_n = -1.

Now compute lim sup⁡bn\limsup b_n and lim inf⁡bn\liminf b_n. For even nn: bn=−1+2/n→−1b_n = -1 + 2/n \to -1. For odd nn: bn=1+2/n→1b_n = 1 + 2/n \to 1. So Sb={1,−1}S_b = \{1,-1\} also, giving

lim sup⁡bn=1,lim inf⁡bn=−1.\limsup b_n = 1, \qquad \liminf b_n = -1.

Now form the sum an+bna_n + b_n. For even nn:

an+bn=(1−1n)+(−1+2n)=1n→0.a_n + b_n = \left(1 - \frac1n\right) + \left(-1 + \frac2n\right) = \frac1n \to 0.

For odd nn:

an+bn=(−1+1n)+(1+2n)=3n→0.a_n + b_n = \left(-1+\frac1n\right) + \left(1+\frac2n\right) = \frac3n \to 0.

So an+bn→0a_n + b_n \to 0 along both parities, hence (since these two classes exhaust all indices) lim⁡n(an+bn)=0\lim_n (a_n+b_n) = 0, and therefore

lim sup⁡(an+bn)=lim inf⁡(an+bn)=0.\limsup(a_n+b_n) = \liminf(a_n+b_n) = 0.

Compare with the general subadditivity/superadditivity inequalities (which follow from Part 3 by a short argument: any subsequential limit of an+bna_n+b_n along a subsequence where both anj→α∈Saa_{n_j}\to\alpha\in S_a and (further refining) bnj→β∈Sbb_{n_j}\to\beta \in S_b satisfies α+β≤La∗+Lb∗\alpha+\beta \le L^*_a + L^*_b, giving lim sup⁡(an+bn)≤lim sup⁡an+lim sup⁡bn\limsup(a_n+b_n)\le \limsup a_n + \limsup b_n, and dually for liminf):

lim sup⁡(an+bn)≤lim sup⁡an+lim sup⁡bn=1+1=2,here 0≤2,\limsup(a_n+b_n) \le \limsup a_n + \limsup b_n = 1 + 1 = 2, \qquad \text{here } 0 \le 2,
lim inf⁡(an+bn)≥lim inf⁡an+lim inf⁡bn=−1+(−1)=−2,here 0≥−2.\liminf(a_n+b_n) \ge \liminf a_n + \liminf b_n = -1 + (-1) = -2, \qquad \text{here } 0 \ge -2.

Both inequalities hold, and both are strict: 0<20 < 2 and 0>−20 > -2. This is because the sequences were tuned so that ana_n's "high" phase (even nn, near 11) coincides with bnb_n's "low" phase (even nn, near −1-1), so the two never achieve their extremes simultaneously — precisely the phenomenon the next counterexample isolates.

CounterexampleWhy lim sup⁡(an+bn)=lim sup⁡an+lim sup⁡bn\limsup(a_n+b_n) = \limsup a_n + \limsup b_n can fail: no hypothesis of monotonicity/convergence of ana_n or bnb_n can be dropped and expect equality in general

The theorem above never claimed lim sup⁡(an+bn)=lim sup⁡an+lim sup⁡bn\limsup(a_n + b_n) = \limsup a_n + \limsup b_n; it only claimed ≤\le. It is instructive to see explicitly why equality is not a theorem — i.e., why one cannot strengthen Part 3's corollary to an equality without an extra hypothesis such as convergence of one of the two sequences. Take, as in Example 2,

an=(−1)n+(−1)n+1n,bn=(−1)n+1+2n.a_n = (-1)^n + \frac{(-1)^{n+1}}{n}, \qquad b_n = (-1)^{n+1} + \frac{2}{n}.

We computed lim sup⁡an=1\limsup a_n = 1, lim sup⁡bn=1\limsup b_n = 1, so the claim "lim sup⁡(an+bn)=lim sup⁡an+lim sup⁡bn\limsup(a_n+b_n) = \limsup a_n+\limsup b_n" would assert

lim sup⁡n(an+bn)2.\limsup_n (a_n + b_n) \stackrel{?}{=} 2.

But we directly computed an+bn→0a_n + b_n \to 0, so lim sup⁡(an+bn)=0≠2\limsup(a_n+b_n) = 0 \ne 2. The equality fails.

Diagnosis. The inequality lim sup⁡(an+bn)≤lim sup⁡an+lim sup⁡bn\limsup(a_n+b_n) \le \limsup a_n + \limsup b_n is proved by bounding, for each nn, sup⁡k≥n(ak+bk)≤sup⁡k≥nak+sup⁡k≥nbk\sup_{k\ge n}(a_k+b_k) \le \sup_{k\ge n} a_k + \sup_{k\ge n} b_k — this only uses that ak≤sup⁡k≥naka_k \le \sup_{k\ge n}a_k and bk≤sup⁡k≥nbkb_k \le \sup_{k\ge n} b_k simultaneously at the same index kk. Equality in sup⁡(f+g)≤sup⁡f+sup⁡g\sup(f+g)\le \sup f + \sup g requires the two suprema to be approached (or attained) at a common (or asymptotically common) point. Here ana_n's supremum-approaching indices are the even nn's, and bnb_n's supremum-approaching indices are the odd nn's — disjoint sets — so the two sequences never simultaneously get close to their respective limsups. This is exactly the missing hypothesis: if one of the two sequences converges (so that its "sup-approaching set" is eventually all indices, not a proper subset), equality does hold: if an→Aa_n \to A (a genuine limit), then

lim sup⁡(an+bn)=A+lim sup⁡bn\limsup(a_n+b_n) = A + \limsup b_n

always, because for the convergent sequence bn′:=anb_n' := a_n, sup⁡k≥n(ak+bk)\sup_{k \ge n}(a_k+b_k) and A+sup⁡k≥nbkA + \sup_{k\ge n} b_k differ by at most sup⁡k≥n∣ak−A∣→0\sup_{k\ge n}|a_k - A| \to 0. Dropping "one of the two sequences converges" — as in our counterexample, where neither ana_n nor bnb_n converges — is precisely what allows strict inequality, and the example shows the gap can be as large as possible relative to the given limsups (here the gap is 2−0=22 - 0 = 2).

Scope, edge cases & common traps

SituationWhat happensWhy
(an)(a_n) unbounded abovelim sup⁡an=+∞\limsup a_n = +\inftytail suprema bn=+∞b_n = +\infty for all nn
(an)(a_n) unbounded belowlim inf⁡an=−∞\liminf a_n = -\inftytail infima cn=−∞c_n = -\infty for all nn
(an)(a_n) convergentlim inf⁡an=lim sup⁡an=lim⁡an\liminf a_n = \limsup a_n = \lim a_nPart 4 of the theorem
(an)(a_n) bounded, not convergentlim inf⁡an<lim sup⁡an\liminf a_n < \limsup a_n, both finiteSS has ≥2\ge 2 points, by Part 3
Only finitely many terms alteredlim sup⁡,lim inf⁡\limsup, \liminf unchangedtails bn,cnb_n, c_n for large nn ignore any finite prefix
an≤bna_n \le b_n for all nn (termwise)lim sup⁡an≤lim sup⁡bn\limsup a_n \le \limsup b_n and lim inf⁡an≤lim inf⁡bn\liminf a_n \le \liminf b_nmonotonicity of sup/inf passes to limits

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · early 19th-century French analysis, rigorization of calculus

Cauchy did not use the modern notation lim sup⁡\limsup/lim inf⁡\liminf, but the underlying idea — bounding the eventual oscillation of a sequence or the terms of a series to test convergence — traces to his work on rigorizing analysis, particularly his root-test criterion for series, later sharpened using the limsup formalism by Jacques Hadamard in 1888 for the radius of convergence of power series. The formal apparatus of limit points, subsequential limits, and the modern lim sup⁡/lim inf⁡\limsup/\liminf notation was consolidated over the 19th century as part of the broader arithmetization of analysis associated with Cauchy, Bolzano, and Weierstrass.

History
- 1821: Cauchy's Cours d'Analyse introduces rigorous ε\varepsilon-style convergence arguments and an early root-test criterion for series, implicitly using bounds on tail behavior.
- 1817: Bolzano gives an early, largely unpublished-in-its-time proof of what becomes the Bolzano–Weierstrass theorem, the structural fact (nonempty compact set of subsequential limits) underlying lim sup⁡\limsup/lim inf⁡\liminf.
- 1860s–1870s: Weierstrass's Berlin lectures develop the modern ε\varepsilon-δ\delta/subsequence framework in which limit points of sequences are systematically studied.
- 1888: Hadamard uses lim sup⁡\limsup of ∣cn∣1/n|c_n|^{1/n} to give the exact formula for the radius of convergence of a power series, cementing the practical importance of limsup beyond simple sequence convergence.
- Late 19th–early 20th century: the notations lim⁡‾\overline{\lim}, lim⁡‾\underline{\lim} (and later lim sup⁡\limsup, lim inf⁡\liminf) become standard in analysis textbooks worldwide.
Do lim sup⁡an\limsup a_n and lim inf⁡an\liminf a_n always exist, even if lim⁡an\lim a_n does not?

Yes. This is exactly Part 1 of the central theorem: the tail suprema bn=sup⁡k≥nakb_n = \sup_{k\ge n}a_k form a nonincreasing sequence in the extended reals [−∞,∞][-\infty,\infty], and every monotone sequence in the extended reals converges there (to a finite limit, or to ±∞\pm\infty). The same holds for the tail infima cnc_n. So lim sup⁡an\limsup a_n and lim inf⁡an\liminf a_n are defined for literally every sequence of real numbers, with values possibly +∞+\infty or −∞-\infty. This is precisely what makes them more useful than lim⁡an\lim a_n in convergence tests: you never have to first check that the ordinary limit exists.

How exactly does lim sup⁡an=lim inf⁡an\limsup a_n = \liminf a_n imply lim⁡an\lim a_n exists?

This is Part 4 of the theorem, proved via the squeeze principle. Since cn≤an≤bnc_n \le a_n \le b_n for every nn (the tail infimum and supremum always sandwich the corresponding term), if both bn→Lb_n \to L and cn→Lc_n \to L for the same extended real LL, then ana_n is squeezed between two sequences converging to LL, forcing an→La_n \to L as well. The squeeze argument works identically whether LL is finite or ±∞\pm\infty, just replacing ε\varepsilon-neighborhoods with the appropriate extended-real neighborhoods.

Is lim sup⁡an\limsup a_n the same as the largest term of the sequence, or the supremum of all the ana_n?

No, and this is a very common confusion. The supremum sup⁡nan\sup_n a_n (over all nn) can be strictly larger than lim sup⁡nan\limsup_n a_n, because sup⁡nan\sup_n a_n can be dominated by a single early outlier term that never recurs. For example, if a1=100a_1 = 100 and an=1/na_n = 1/n for all n≥2n \ge 2, then sup⁡nan=100\sup_n a_n = 100, but lim sup⁡nan=0\limsup_n a_n = 0, since the tail suprema bn=sup⁡k≥nakb_n = \sup_{k\ge n} a_k eventually ignore the term a1a_1 once n≥2n \ge 2, and bn=1/n→0b_n = 1/n \to 0. The limsup only cares about the eventual, tail behavior, not any finite prefix.

What is the relationship between lim sup⁡an\limsup a_n and the set of subsequential limits?

By Part 3 of the central theorem, if SS denotes the set of all subsequential limits of (an)(a_n) in the extended reals, then SS is always nonempty, and lim sup⁡an=max⁡S\limsup a_n = \max S while lim inf⁡an=min⁡S\liminf a_n = \min S. This is often the most useful mental model: to find lim sup⁡an\limsup a_n, find every possible value a subsequence of (an)(a_n) could converge to (including ±∞\pm\infty), and take the largest one. This is exactly how Example 1 in this article was computed, by splitting the sequence into residue classes and finding each class's subsequential limit(s).

Why does lim sup⁡(an+bn)≤lim sup⁡an+lim sup⁡bn\limsup(a_n+b_n) \le \limsup a_n + \limsup b_n sometimes fail to be an equality?

The inequality itself always holds (it follows from sup⁡k≥n(ak+bk)≤sup⁡k≥nak+sup⁡k≥nbk\sup_{k\ge n}(a_k+b_k) \le \sup_{k\ge n}a_k + \sup_{k\ge n}b_k, since each ak+bka_k+b_k is bounded above by the sum of the two separate tail suprema). But equality can fail because the two sequences might approach their respective supremum-behavior along different subsequences of indices, so they are never simultaneously "near their peak." The counterexample in this article, with ana_n peaking on even nn and bnb_n peaking on odd nn, shows the gap can be as large as the entire difference between lim sup⁡an+lim sup⁡bn\limsup a_n+\limsup b_n and the true value. Equality is guaranteed if one of the two sequences actually converges (has an honest limit).

How are lim sup⁡\limsup and lim inf⁡\liminf used in the Root Test and radius of convergence?

Both the Root Test and the Cauchy–Hadamard formula for the Radius of Convergence rely on lim sup⁡\limsup precisely because the relevant ordinary limits (lim⁡∣an∣1/n\lim |a_n|^{1/n}, or lim⁡∣cn+1/cn∣\lim |c_{n+1}/c_n|) frequently fail to exist even for perfectly well-behaved, convergent series or power series. The Root Test states ∑an\sum a_n converges absolutely if lim sup⁡n∣an∣1/n<1\limsup_n |a_n|^{1/n} < 1; because lim sup⁡\limsup always exists, this test applies universally, unlike a naive version using lim⁡\lim. Similarly R=1/lim sup⁡n∣cn∣1/nR = 1/\limsup_n |c_n|^{1/n} always makes sense and gives the exact radius of convergence, with no assumption that the ordinary limit of ∣cn∣1/n|c_n|^{1/n} exists.

Can lim sup⁡an\limsup a_n or lim inf⁡an\liminf a_n equal +∞+\infty or −∞-\infty, and what does that mean intuitively?

Yes, whenever the sequence is unbounded in the relevant direction. lim sup⁡an=+∞\limsup a_n = +\infty means the tail suprema sup⁡k≥nak\sup_{k\ge n}a_k never become finite, i.e. arbitrarily large values of ana_n recur no matter how far out you look; equivalently (by Part 3) some subsequence diverges to +∞+\infty. Symmetrically lim inf⁡an=−∞\liminf a_n = -\infty means arbitrarily negative values recur indefinitely. These are legitimate, well-defined extended-real values, not failures of the definition — this is exactly the sense in which limsup/liminf are defined for literally every sequence, unlike the ordinary limit.

Before this

Where this leads