Alphabeta Math
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 19 results · all verified · 19 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 19 also cleared it.

limsup, liminf, and Subsequential Limits

1 · Prerequisites

2 · Summary

Objective. Every real sequence has a largest and a smallest thing it keeps coming back to. This page defines those two quantities, proves that they always exist, and shows that their coincidence is exactly convergence. The obstacle is that neither quantity is a real number in general, and the previous pages deliberately refused to pretend otherwise: Conventions: sup⁡∅, unbounded sets, and the extended reals barred the conventions sup⁡S=+∞ and inf⁡∅=+∞ inside R, and promised that a page needing the extended line would introduce it explicitly as a new object rather than quietly enlarging R. This is that page, and the promise is discharged in its first two items.

The extended line, built once and used everywhere. The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined adjoins two objects −∞ and +∞ to R, fixes a total order in which they are the least and greatest elements, and defines exactly two partial operations, leaving (+∞)+(−∞) and 0⋅(±∞) undefined. Nothing about R is changed, and no algebraic law is inherited: R‾ is not a field. What it does have is Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R: every subset of R‾, with no hypothesis whatever, has a least upper bound and a greatest lower bound there, agreeing with the real supremum and infimum wherever the latter are defined. That single lemma is what makes the whole page hypothesis free, and fourteen of its items rest on it.

The two quantities. Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾ sets lim sup⁡kxk=inf⁡nsup⁡k≥nxk and lim inf⁡kxk=sup⁡ninf⁡k≥nxk, both taken in R‾; The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence records that the tail suprema decrease and the tail infima increase, so the outer operations act on monotone families, and that both quantities exist for every sequence, bounded or not. lim sup⁡(−xk)=−lim inf⁡(xk), with the reflection of R‾ exchanging ±∞ proves that x↦−x exchanges the two, which is what lets every later statement about lim sup⁡ be read off as a statement about lim inf⁡ without a second proof. lim inf⁡xk≤lim sup⁡xk for every real sequence puts them in order, and For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently gives the working form in the finite case: L=lim sup⁡kxk exactly when xk is eventually below L+ε and frequently above L−ε, for every ε>0. The asymmetry between eventually and frequently is the whole content of the notion.

What they are for. A real sequence converges to L∈R iff lim inf⁡xk=lim sup⁡xk=L, and diverges to ±∞ iff both equal ±∞ is the theorem that justifies the definitions: a sequence converges to a real L exactly when both quantities equal L, and diverges to ±∞ exactly when both equal ±∞. So a question about convergence becomes a question about two quantities that always exist, and a proof of convergence can be assembled from one-sided estimates with no candidate limit in hand. The limit superior is itself a subsequential limit in R‾ and is the greatest one identifies them intrinsically: lim sup⁡kxk is itself a subsequential limit and is the greatest one, with The limit inferior is the least subsequential limit in R‾ the dual. Both statements live in R‾, and they have to: the greatest subsequential limit of an unbounded sequence need not be real, and the finite subsequential limit set may have a greatest element that is not the limit superior. Since the published Subsequential limit of a real sequence, and the subsequential limit set is finite by design, Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞ extends it by citation, adding the two divergence clauses without touching what was already fixed. If each yj is a subsequential limit of (xk) and yj→y∈R, then y is a subsequential limit of (xk) completes the picture from the other side: the subsequential limit set contains the limit of every convergent sequence of its own points.

Calculus of the two quantities. If xk≤yk eventually then lim sup⁡xk≤lim sup⁡yk and lim inf⁡xk≤lim inf⁡yk is the comparison principle, lim sup⁡(xk+yk)≤lim sup⁡xk+lim sup⁡yk whenever the right-hand side is defined in R‾, and dually for lim inf⁡ the subadditivity, with the hypothesis that the right-hand side be defined in R‾ and not a hypothesis more, and For bounded nonnegative sequences, lim sup⁡(xkyk)≤(lim sup⁡xk)(lim sup⁡yk) the multiplicative analogue for bounded nonnegative sequences, where both hypotheses are load bearing. Neither inequality can be reversed: FALSE: lim sup⁡(xk+yk)=lim sup⁡xk+lim sup⁡yk records that for the sum, and xk=(−1)k, yk=(−1)k+1 give lim sup⁡(xk+yk)=0<2=lim sup⁡xk+lim sup⁡yk and xk=1+(−1)k, yk=1+(−1)k+1 give lim sup⁡(xkyk)=0<4 exhibit strict inequality in each case.

The ratio-to-root chain, and the standard limits. For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak proves lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak for a positive sequence. This is the precise sense in which a root criterion dominates a ratio criterion, and it belongs here rather than with series, because it is a statement about lim sup⁡ and nothing else. FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak for every positive sequence shows the outer inequalities are not equalities. The proof needs For every a>0, a1/n→1, which with n1/n→1, For every p>0 and every positive rational α, nα/(1+p)n→0 and For every real x, xk/k!→0 makes up the four standard limits of elementary analysis; the last two are proved directly from Bernoulli's inequality and the Archimedean property, since continuity of x↦xα is not available at this point in the reading order; it is proved later in Continuity and derivatives of positive-base real powers.

A note on indices. Sequences here are functions on N and N contains 0 (Sequences of reals: bounded, eventually, frequently, tails, subsequences). The expressions n1/n, a1/n and ak1/k are undefined at index 0, so the corresponding statements are made about the shifted families (k+1)1/(k+1), a1/(k+1) and ak+11/(k+1); the expressions nα/(1+p)n and xk/k! are defined at 0 and are not shifted. Each item says which case it is in.

What is left undefined, and where it bites. Which extended-real operations this library leaves undefined, and where each lim sup⁡ statement needs the hypothesis collects the two gaps in the arithmetic of R‾, says why no value could fill them, and goes through the page recording which statements carry a hypothesis because of them and which are unconditional. The short version: everything order-theoretic is hypothesis free, and everything arithmetic is not.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined

Definition

Fix two objects −∞ and +∞, distinct from one another and neither of them a real number (The real numbers), and set

R‾:=R∪{−∞,+∞}.

This is a new object, introduced here explicitly with its own order and its own partial arithmetic. It is not an enlargement of the field R, and no operation of R (Complete ordered field (least-upper-bound property)) is redefined by anything below.

The order. For a,b∈R‾ declare

a≤b:⟺a=−∞  or  b=+∞  or  (a,b∈R and a≤b in R),

with R ordered as in Order on the reals, and write a<b for "a≤b and a≠b" as usual (Partial order and partially ordered set).

(R‾,≤) is a totally ordered set, and the inclusion of R preserves and reflects the order. All four checks are immediate from the displayed clauses.

  • Reflexive. For a=±∞ one of the first two clauses applies; for a∈R the third does, since a≤a in R.
  • Antisymmetric. Suppose a≤b and b≤a. If a=−∞ then b≤a forces b=−∞, since the clause a=+∞ fails and b,a are not both real. Symmetrically b=−∞ forces a=−∞, and a=+∞ or b=+∞ forces the other to be +∞. In the one remaining situation a and b are both real and antisymmetry is that of R.
  • Transitive. Let a≤b≤c. If a=−∞ or c=+∞ the conclusion is one of the first two clauses. Otherwise a≠−∞ forces, in a≤b, either b=+∞ or a,b∈R; and c≠+∞ forces, in b≤c, either b=−∞ or b,c∈R. The value b=+∞ is incompatible with the second alternative pair, so b is real, hence so are a and c, and transitivity is that of R.
  • Total. If a=−∞ or b=+∞ then a≤b; if b=−∞ or a=+∞ then b≤a; otherwise both are real and the order of R is total.
  • Preserved and reflected. For a,b∈R the first two clauses fail, so a≤b in R‾ says exactly a≤b in R.

In particular −∞ is the least and +∞ the greatest element of R‾, and −∞<x<+∞ for every x∈R.

Reflection. Extend negation by

−(+∞):=−∞,−(−∞):=+∞,

keeping the field negative on R. The resulting map ν:R‾→R‾, ν(a)=−a, satisfies ν(ν(a))=a and

a≤b  ⟺  −b≤−a(a,b∈R‾).

For a and b real this is the elementwise order reversal in R: translation invariance (Order is preserved by adding a constant and by adding inequalities) applied with the constant −a−b turns a<b into −b<−a and, applied with the constant a+b, turns it back, while a=b holds exactly when −a=−b. In every other case both sides are decided by the first two clauses of the order: a=−∞ makes both sides true, as does b=+∞, and if a≠−∞, b≠+∞ and a,b are not both real then one of a=+∞, b=−∞ holds and both sides are false.

Partial addition. For a,b∈R‾ the sum a+b is defined by

  • a+b = the field sum, when a,b∈R;
  • a+b:=+∞ when a=+∞ and b≠−∞, or b=+∞ and a≠−∞;
  • a+b:=−∞ when a=−∞ and b≠+∞, or b=−∞ and a≠+∞;

and the two sums (+∞)+(−∞) and (−∞)+(+∞) are left undefined. Addition is commutative where defined, and

−(a+b)=(−a)+(−b),

each side being defined exactly when the other is: the excluded pairs {+∞,−∞} are exchanged by ν, and the three clauses above are exchanged accordingly.

Partial multiplication. For a,b∈R‾ the product ab is defined by

  • ab = the field product, when a,b∈R;
  • ab:=+∞ when one of a,b is ±∞, the other is ≠0, and both are >0 or both are <0;
  • ab:=−∞ when one of a,b is ±∞, the other is ≠0, and one is >0 and the other <0;

and every product with one factor 0 and the other ±∞ is left undefined. The comparisons >0 and <0 here are taken in the order above, under which +∞>0>−∞.

Nothing else is defined. There is no subtraction, no division, no exponentiation and no absolute value on R‾ in this library; where such an expression is wanted it is written out in the two defined operations, and where a case falls in the undefined list the statement carries an explicit hypothesis saying so.

Remarks

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R

Statement

Let A⊆R‾ be any subset of the extended real line (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined) and write AR:=A∩R. Then A has a least upper bound and a greatest lower bound in R‾ (Upper bound, least upper bound, and strict upper bound), each unique, which we write sup⁡A and inf⁡A with the ambient set always R‾. Explicitly:

  • sup⁡A=+∞ if +∞∈A, or if AR is not bounded above in R;
  • sup⁡A=−∞ if +∞∉A and AR=∅;
  • sup⁡A is the real supremum sup⁡AR (Complete ordered field (least-upper-bound property)) if +∞∉A and AR is nonempty and bounded above in R;

and dually, with −∞ and +∞ exchanged and "above" replaced by "below", for inf⁡A (Greatest lower bound (infimum), Every nonempty set bounded below has an infimum).

Agreement. If A⊆R is nonempty and bounded above in R (Lower bound, bounded below, bounded set) then sup⁡A computed in R‾ is the real number sup⁡A of Complete ordered field (least-upper-bound property); if A⊆R is nonempty and bounded below then inf⁡A computed in R‾ is the real number inf⁡A of Every nonempty set bounded below has an infimum. In particular the notation is unambiguous on the sets for which the real supremum and infimum are defined, and sup⁡∅=−∞, inf⁡∅=+∞ in R‾.

No hypothesis is placed on A. This is exactly what the real supremum cannot do, and it is why every lim sup⁡ statement on this page holds for every sequence rather than for bounded ones only. It is also not a weakening of the discipline this library keeps around suprema: the operation supplied here is a different operation, taken in a different ordered set, and the agreement clause records exactly where the two coincide.

Facts & Assumptions

Given: A subset A⊆R‾, and its real part AR:=A∩R.

[L1]

(R‾,≤) is a totally ordered set in which −∞ is the least element and +∞ the greatest, and whose order restricted to R is the order of R (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set, Order on the reals).

[L2]

Upper and lower bounds in a poset: u is an upper bound of A when a≤u for all a∈A, and a least upper bound when moreover u≤v for every upper bound v; dually for lower bounds and greatest lower bounds. Each is unique when it exists, by antisymmetry (Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L3]

Least-upper-bound property of R: every nonempty S⊆R that is bounded above in R has a real least upper bound sup⁡S (Complete ordered field (least-upper-bound property)).

[L4]

Greatest-lower-bound property of R: every nonempty S⊆R that is bounded below in R has a real greatest lower bound inf⁡S (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

[L5]

Bounded above and bounded below in R mean the existence of a real upper, respectively lower, bound (Lower bound, bounded below, bounded set).

Proof

technique · cases
1.1

Case S1 for the supremum: +∞∈A.

givenassume-case suptop
1.2

Case S2 for the supremum: +∞∉A and AR=∅, so that every element of A equals −∞.

givenassume-case supbot
1.3

Case S3 for the supremum: +∞∉A, AR≠∅, and AR is bounded above in R.

givenassume-case supfin
1.4

Case S4 for the supremum: +∞∉A, AR≠∅, and AR is not bounded above in R.

givenassume-case supunb
1.5

Case I1 for the infimum: −∞∈A.

givenassume-case infbot
1.6

Case I2 for the infimum: −∞∉A and AR=∅, so that every element of A equals +∞.

givenassume-case inftop
1.7

Case I3 for the infimum: −∞∉A, AR≠∅, and AR is bounded below in R.

givenassume-case inffin
1.8

Case I4 for the infimum: −∞∉A, AR≠∅, and AR is not bounded below in R.

givenassume-case infunb
2.1

In case S1 the element +∞ is an upper bound of A, being the greatest element of R‾; and if v is any upper bound of A then +∞∈A gives +∞≤v, whence v=+∞ by antisymmetry. So +∞ is the least upper bound of A.

step 1.1L1L2
2.2

In case S2 every element of A equals −∞, so −∞ is an upper bound of A by reflexivity; and −∞≤v for every v∈R‾, being the least element. So −∞ is the least upper bound of A.

step 1.2L1L2
2.3

In case S3 the real number σ:=sup⁡AR exists, and it is an upper bound of A in R‾: an element of A is either real, hence lies in AR and satisfies a≤σ in R and so in R‾, or equals −∞, which is ≤σ; the value +∞ does not occur in A in this case.

step 1.3L1L3
2.4

In case S4 the element +∞ is an upper bound of A; and if v is an upper bound then v≠−∞, because fixing a∈AR, which is possible in this case, gives a≤v with a real and −∞ is below no real, while v real would make v a real upper bound of AR and contradict the case hypothesis. So v=+∞, and +∞ is the least upper bound of A.

step 1.4L1L2L5
2.5

In case I1 the element −∞ is a lower bound of A, being least; and any lower bound w satisfies w≤−∞ because −∞∈A, whence w=−∞ by antisymmetry. So −∞ is the greatest lower bound of A.

step 1.5L1L2
2.6

In case I2 every element of A equals +∞, so +∞ is a lower bound of A by reflexivity, and w≤+∞ for every w. So +∞ is the greatest lower bound of A.

step 1.6L1L2
2.7

In case I3 the real number ι:=inf⁡AR exists and is a lower bound of A in R‾: an element of A is either real, hence in AR and ≥ι, or equals +∞≥ι; the value −∞ does not occur in A in this case.

step 1.7L1L4
2.8

In case I4 the element −∞ is a lower bound of A; any lower bound w satisfies w≠+∞, because fixing a∈AR gives w≤a with a real and +∞ is above no real, while w real would be a real lower bound of AR and contradict the case hypothesis. So w=−∞ is the greatest lower bound of A.

step 1.8L1L2L5
3.1

In case S3 let v be any upper bound of A and fix a∈AR, which is possible since AR≠∅. From a≤v with a real we get v≠−∞, since −∞ is below no real. If v=+∞ then σ≤v because +∞ is greatest. Otherwise v is real, and it bounds AR above in R, so σ≤v by leastness of the real supremum. Hence σ is the least upper bound of A.

step 1.3step 2.3L1L2L3
3.2

In case I3 let w be a lower bound of A and fix a∈AR. From w≤a with a real we get w≠+∞. If w=−∞ then w≤ι; otherwise w is real and bounds AR below in R, so w≤ι. Hence ι is the greatest lower bound of A.

step 1.7step 2.7L1L2L4
4.1

The four supremum cases are exhaustive and mutually exclusive: either +∞∈A, which is S1, or not, and then either AR=∅, which is S2, or AR≠∅ and it is bounded above in R, which is S3, or it is not, which is S4. In each case a least upper bound was produced, and it is unique. The same four alternatives with −∞, +∞ and "below" in place of +∞, −∞ and "above" are I1 to I4, and in each a greatest lower bound was produced.

step 2.1step 2.2step 3.1step 2.4step 2.5step 2.6step 3.2step 2.8L2L5cases: a two-fold split followed by a three-fold splitcases-exhaustive
5.1

The agreement clause follows: a nonempty A⊆R bounded above in R satisfies +∞∉A and AR=A, so case S3 applies and sup⁡A=sup⁡AR is the real supremum; a nonempty A⊆R bounded below satisfies case I3 and inf⁡A is the real infimum; and A=∅ falls under S2 and I2, giving sup⁡∅=−∞ and inf⁡∅=+∞.

step 2.3step 3.1step 2.7step 3.2step 4.1L3L4∎

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞

Definition

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let L∈R‾ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined). Say that (xk) converges to L in R‾ when one of the following holds, according to which of the three kinds of element L is:

Then L is an extended subsequential limit of (xk) when some subsequence of (xk) converges to L in R‾: when there is a strictly increasing n:N→N (Sequences of reals: bounded, eventually, frequently, tails, subsequences) such that (xnj)j∈N converges to L in the sense just given. The extended subsequential limit set of (xk) is

SL⁡‾(x)  :=  { L∈R‾:L is an extended subsequential limit of (xk) }⊆R‾.

This extends the published Subsequential limit of a real sequence, and the subsequential limit set and does not replace it. That definition is finite by design: there L ranges over R and SL⁡(x)⊆R. Its clause is quoted verbatim as the first of the three clauses above, so

SL⁡‾(x)∩R=SL⁡(x),

immediately from the definitions: a real L lies in SL⁡‾(x) exactly when some subsequence converges to L in the sense of Limits and Cauchy sequences of reals, which is exactly the condition L∈SL⁡(x). The extended set is therefore SL⁡(x) together with at most the two extra points ±∞, each present exactly when some subsequence diverges to it. Nothing about SL⁡(x) is redefined, and every statement proved about SL⁡(x) elsewhere in the library remains a statement about the same set.

Neither is Divergence to +∞ and to −∞ reinterpreted. The phrase "xk→+∞" keeps exactly the meaning fixed there, an abbreviation for "for every real M, eventually xk>M". What is new is only that the phrase is now allowed to appear as one of three clauses in a single definition whose parameter L ranges over R‾, so that the three situations can be quantified over together. In particular the warning recorded there stands: a sequence diverging to +∞ has no limit in R, and none of the rules of Algebra of limits: sums, scalar multiples, products and quotients applies to it.

Remarks

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾

Definition

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). For n∈N let

Tn  :=  { xk:k∈N, k≥n }⊆R

be the n-th tail range of (xk), a nonempty subset of R since xn∈Tn. Regard Tn as a subset of R‾ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined) and put

sn  :=  sup⁡Tn∈R‾,in  :=  inf⁡Tn∈R‾,

the supremum and infimum taken in R‾, which exist for every n and for every sequence by Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R. The limit superior and limit inferior of (xk) are then

lim sup⁡kxk  :=  inf⁡{ sn:n∈N },lim inf⁡kxk  :=  sup⁡{ in:n∈N },

again taken in R‾ and again existing by Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R, since {sn:n∈N} and {in:n∈N} are subsets of R‾ on which no hypothesis is needed. Both are elements of R‾, and either may be +∞ or −∞. The notations lim sup⁡k→∞xk, lim‾⁡kxk and lim⁡‾kxk all denote the first of them elsewhere; this library writes lim sup⁡kxk.

Every quantity written here exists, and that is why the extended line was introduced. Each of the four operations above is an application of Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R to a subset of R‾ carrying no hypothesis whatever. Written with the real supremum of Complete ordered field (least-upper-bound property) and the real infimum of Every nonempty set bounded below has an infimum instead, the definition would be available only for sequences that are bounded (Lower bound, bounded below, bounded set): sup⁡Tn needs Tn bounded above, and inf⁡{sn} needs {sn} nonempty, bounded below, and made of real numbers (Greatest lower bound (infimum)). None of those is automatic, and the discipline recorded in Conventions: sup⁡∅, unbounded sets, and the extended reals forbids papering over the gap with a convention. The extended supremum is a different operation in a different ordered set, and it is total.

Values, when the sequence is bounded. If (xk) is bounded, say ∣xk∣≤M for every k, then each Tn is a nonempty subset of R bounded above by M and below by −M, so by the agreement clause of Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R each sn and each in is the real supremum or infimum of Tn, and lies in [−M,M]. The family {sn} is then a nonempty set of reals bounded below by −M, so lim sup⁡kxk is likewise the real infimum of {sn} and lies in [−M,M]; dually for lim inf⁡kxk. So for a bounded sequence both quantities are ordinary real numbers computed with the ordinary real supremum and infimum, and the extended line is doing no work. It is only for unbounded sequences that the values ±∞ occur.

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), with tail ranges Tn and extended tail bounds sn=sup⁡Tn, in=inf⁡Tn as in Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾.

  1. Monotonicity of the extended bounds under inclusion. If A⊆B⊆R‾ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined) then sup⁡A≤sup⁡Bandinf⁡B≤inf⁡A, the four quantities being the extended bounds of Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R. No hypothesis is placed on A or B; in particular A may be empty.
  2. The tail bounds are monotone. Tm⊆Tn whenever n≤m, and hence sm≤snandin≤im(n≤m). In particular sn+1≤sn and in≤in+1 for every n, and in≤sn for every n.
  3. Existence. lim sup⁡kxk and lim inf⁡kxk exist in R‾ for every sequence of reals, bounded or not.

Claim 1 is the tool the rest of this page uses whenever two extended suprema are compared. It is proved here, from the definition of a least upper bound, rather than quoted from the suprema page, for the reason given in the remarks below.

Facts & Assumptions

Given: A sequence (xk) of reals, its tail ranges Tn={xk:k≥n}, and the extended bounds sn=sup⁡Tn, in=inf⁡Tn (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L1]

Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, with no hypothesis on the subset (Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R).

[L2]

Least upper bound and greatest lower bound in a poset: sup⁡A is an upper bound of A that is ≤ every upper bound of A, and inf⁡A is a lower bound that is ≥ every lower bound; each is unique when it exists (Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L4]

The order on N is total and transitive (Order on the natural numbers, ≤ is a linear order on N).

Proof

technique · direct
1.1

Let A⊆B⊆R‾ be arbitrary. By [L1] the four elements sup⁡A, sup⁡B, inf⁡A, inf⁡B of R‾ all exist and are uniquely determined.

givenL1L2
1.2

Let n≤m in N. Every element of Tm has the form xk with k≥m, and then k≥n by transitivity, so xk∈Tn; hence Tm⊆Tn.

givenL4
1.3

For every n the tail range Tn contains xn, so in≤xn because in is a lower bound of Tn, and xn≤sn because sn is an upper bound of Tn; transitivity gives in≤sn.

givenL1L2L3
2.1

Since sup⁡B is an upper bound of B and A⊆B, every element of A is ≤sup⁡B, so sup⁡B is an upper bound of A; as sup⁡A is the least of the upper bounds of A, this gives sup⁡A≤sup⁡B. Dually inf⁡B is a lower bound of B, hence of A, and as inf⁡A is the greatest of the lower bounds of A this gives inf⁡B≤inf⁡A. Claim 1 is proved.

step 1.1L1L2
3.1

Applying claim 1 to the inclusion Tm⊆Tn valid for n≤m gives sm≤sn and in≤im; the special case m=n+1 gives sn+1≤sn and in≤in+1. Together with in≤sn this is claim 2.

step 1.2step 1.3step 2.1
4.1

The families {sn:n∈N} and {in:n∈N} are subsets of R‾, so [L1] applies to them with no hypothesis, and lim sup⁡kxk=inf⁡{sn} and lim inf⁡kxk=sup⁡{in} exist in R‾ for every sequence of reals. This is claim 3.

step 3.1L1L2∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

lim sup⁡(−xk)=−lim inf⁡(xk), with the reflection of R‾ exchanging ±∞

Statement

Write −A:={−a:a∈A} for A⊆R‾, with the reflection of The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, which fixes no point of {−∞,+∞} but exchanges the two.

  1. Reflection exchanges the extended bounds. For every A⊆R‾, sup⁡(−A)=−inf⁡Aandinf⁡(−A)=−sup⁡A, with the bounds of Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R and no hypothesis on A.
  2. Reflection exchanges lim sup⁡ and lim inf⁡. For every sequence (xk) of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), lim sup⁡k(−xk)=−lim inf⁡kxkandlim inf⁡k(−xk)=−lim sup⁡kxk, with lim sup⁡ and lim inf⁡ as in Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾.

Claim 2 is what turns every statement about lim sup⁡ on this page into its dual about lim inf⁡ without a second proof, exactly as the identity inf⁡S=−sup⁡(−S) does in R. The novelty is only that the reflection now has to move the two new points, and it does: −(+∞)=−∞.

Facts & Assumptions

Given: A sequence (xk) of reals, the reflected sequence yk:=−xk, and for A⊆R‾ the reflected set −A={−a:a∈A}.

[L1]

Reflection on R‾: the map a↦−a satisfies −(−a)=a and a≤b if and only if −b≤−a, for all a,b∈R‾ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

[L2]

Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, with no hypothesis on the subset (Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R).

[L3]

Least upper bound and greatest lower bound in a poset, and their uniqueness (Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L4]

Tail ranges Tn={xk:k≥n}, the extended tail bounds sn=sup⁡Tn and in=inf⁡Tn, and lim sup⁡kxk=inf⁡{sn}, lim inf⁡kxk=sup⁡{in} (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

Proof

technique · direct
1.1

Let A⊆R‾ be arbitrary. Since −(−a)=a for every a, the map a↦−a carries A onto −A and −A onto A, so −(−A)=A; and by [L2] each of sup⁡A, inf⁡A, sup⁡(−A), inf⁡(−A) exists.

givenL1L2
1.2

Let Tn and Tn′ be the tail ranges of (xk) and of (yk)=(−xk). Since yk=−xk, the set Tn′={yk:k≥n} is exactly −Tn.

givenL4
2.1

The element −inf⁡A is an upper bound of −A: for a∈A we have inf⁡A≤a, hence −a≤−inf⁡A by [L1], and every element of −A is such a −a. If v is any upper bound of −A, then for a∈A we get −a≤v, hence −v≤a by [L1], so −v is a lower bound of A and therefore −v≤inf⁡A, which gives −inf⁡A≤v by [L1] again. So −inf⁡A is the least upper bound of −A, that is sup⁡(−A)=−inf⁡A.

step 1.1L1L2L3
3.1

Applying the identity just proved to the set −A in place of A, and using −(−A)=A, gives sup⁡A=−inf⁡(−A); reflecting both sides and using −(−a)=a yields inf⁡(−A)=−sup⁡A. Claim 1 is proved.

step 2.1step 1.1L1
4.1

By claim 1 applied to Tn, the n-th tail supremum of (yk) is sup⁡Tn′=sup⁡(−Tn)=−in, and its n-th tail infimum is inf⁡(−Tn)=−sn.

step 1.2step 2.1step 3.1L4
5.1

Hence the family of tail suprema of (yk) is {−in:n∈N}=−{in:n∈N}, so claim 1 applied to {in} gives lim sup⁡k(−xk)=inf⁡(−{in})=−sup⁡{in}=−lim inf⁡kxk.

step 4.1step 3.1L4L5
6.1

The same identity applied to the sequence (yk), whose reflection is (−yk)=(xk) by [L1], reads lim sup⁡kxk=−lim inf⁡k(−xk); reflecting both sides gives lim inf⁡k(−xk)=−lim sup⁡kxk. Both parts of claim 2 are proved.

step 5.1L1∎

Remarks

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

lim inf⁡xk≤lim sup⁡xk for every real sequence

Statement

For every sequence (xk) of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences),

lim inf⁡kxk  ≤  lim sup⁡kxk

in R‾ (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined). No hypothesis is placed on (xk): both sides exist for every sequence (The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence) and the inequality holds between them in every case, including those in which one or both sides are ±∞.

Facts & Assumptions

Given: A sequence (xk) of reals, its tail ranges Tn={xk:k≥n}, and the extended tail bounds sn=sup⁡Tn, in=inf⁡Tn (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L1]

Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, an upper bound below every upper bound and a lower bound above every lower bound respectively (Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R, Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

Monotonicity of the tail bounds: sm≤sn and in≤im whenever n≤m, and in≤sn for every n; both lim sup⁡kxk=inf⁡{sn} and lim inf⁡kxk=sup⁡{in} exist (The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence, Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

Proof

technique · direct
1.1

Let m,n∈N be arbitrary. The order on N is total, so either m≤n or n≤m; let p be whichever of m and n is the larger, so that m≤p and n≤p.

givenL3choose
2.1

Monotonicity of the tail bounds gives im≤ip and sp≤sn, and ip≤sp holds because Tp is nonempty; chaining these by transitivity yields im≤sn. As m and n were arbitrary, every tail infimum is below every tail supremum.

step 1.1L2L4
3.1

Fix n∈N. By step 2.1 the element sn is an upper bound of the family {im:m∈N}, and lim inf⁡kxk is its least upper bound, so lim inf⁡kxk≤sn.

step 2.1L1L2
4.1

Since n was arbitrary, lim inf⁡kxk is a lower bound of the family {sn:n∈N}, and lim sup⁡kxk is its greatest lower bound, so lim inf⁡kxk≤lim sup⁡kxk.

step 3.1L1L2∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let L∈R, with eventually and frequently as in Sequences of reals: bounded, eventually, frequently, tails, subsequences and lim sup⁡, lim inf⁡ as in Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾.

  1. L=lim sup⁡kxk if and only if for every real ε>0 xk<L+ε  eventuallyandxk>L−ε  frequently.
  2. Dually, L=lim inf⁡kxk if and only if for every real ε>0 xk>L−ε  eventuallyandxk<L+ε  frequently.

The hypothesis L∈R is not a restriction that can be lifted. Both conditions are stated with real ε and real L±ε, so neither has a reading at L=±∞; the infinite cases are handled instead by the convergence theorem later on this page. What the lemma does say is that whenever lim sup⁡kxk happens to be a real number, it is pinned down by the familiar two-sided test: nothing exceeds it by a fixed positive amount from some index on, and something comes within any fixed positive amount of it arbitrarily late.

Facts & Assumptions

Given: A sequence (xk) of reals, a real number L, the tail ranges Tn={xk:k≥n}, the extended tail suprema sn=sup⁡Tn, and Λ:=lim sup⁡kxk=inf⁡{sn:n∈N} (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L2]

The order on R‾ is total, so the failure of a≤b is b<a; it restricts on R to the order of R; and every real number is <+∞ and >−∞ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

A property P of indices holds eventually when it holds for all k≥K for some K, and frequently when for every K it holds for some k≥K (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L5]

Order arithmetic in R: for ε>0 one has L−ε<L<L+ε, and a<b if and only if −b<−a, both by translation invariance; the order is total, so exactly one of a<b, a=b, b<a holds and a<a is impossible (Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

[L6]

Reflection exchanges the two quantities: lim sup⁡k(−xk)=−lim inf⁡kxk and lim inf⁡k(−xk)=−lim sup⁡kxk (lim sup⁡(−xk)=−lim inf⁡(xk), with the reflection of R‾ exchanging ±∞).

Proof

technique · direct
1.1

For the forward implication of claim 1, assume L=Λ and let ε>0 be an arbitrary real.

assume-hypL1
1.2

For the converse implication of claim 1, assume that for every real ε>0 the sequence satisfies xk<L+ε eventually and xk>L−ε frequently.

assume-hypL3
2.1

Under the assumption of step 1.1, L+ε>L=Λ, so L+ε is not a lower bound of {sn}, since Λ is the greatest lower bound; by totality there is n with sn<L+ε. For every k≥n we have xk≤sn, hence xk<L+ε; so xk<L+ε eventually.

step 1.1L1L2L3L5
2.2

Under the assumption of step 1.1, fix n∈N. Then Λ≤sn because Λ is a lower bound of {sn}, and L−ε<L=Λ, so L−ε<sn. Hence L−ε is not an upper bound of Tn, for an upper bound u of Tn satisfies sn≤u; by totality of the order on R there is therefore k≥n with xk>L−ε. As n was arbitrary, xk>L−ε frequently.

step 1.1L1L2L3L5
2.3

Under the assumption of step 1.2, let ε>0 be a real and take N with xk<L+ε for all k≥N. Then L+ε is an upper bound of TN, so sN≤L+ε by leastness, and Λ≤sN because Λ is a lower bound of {sn}; hence Λ≤L+ε.

step 1.2L1L2L3
2.4

Under the assumption of step 1.2, let ε>0 be a real and fix n. There is k≥n with xk>L−ε, and xk≤sn, so L−ε<sn and in particular L−ε≤sn. As n was arbitrary, L−ε is a lower bound of {sn}, so L−ε≤Λ by greatest-lower-boundedness.

step 1.2L1L2L3
3.1

Taking ε=1 in steps 2.3 and 2.4 gives L−1≤Λ≤L+1 with L±1 real, so Λ is neither +∞ nor −∞ and is therefore a real number. Suppose Λ>L and put δ:=Λ−L>0; choosing a natural m≥1 with 1/m<δ and applying step 2.3 with ε=1/m gives Λ≤L+1/m<L+δ=Λ, which is impossible. Suppose instead Λ<L and put δ:=L−Λ>0; choosing m≥1 with 1/m<δ and applying step 2.4 with ε=1/m gives L−1/m≤Λ, that is δ=L−Λ≤1/m<δ, again impossible. By trichotomy Λ=L.

step 2.3step 2.4L2L4L5
4.1

Steps 2.1 and 2.2 prove the forward implication of claim 1 and step 3.1 proves its converse, so claim 1 holds.

step 2.1step 2.2step 3.1
5.1

For claim 2, note that L=lim inf⁡kxk holds exactly when −L=−lim inf⁡kxk=lim sup⁡k(−xk), since negation is injective on R‾. Applying claim 1 to the sequence (−xk) and the real number −L, that holds exactly when for every real ε>0 one has −xk<−L+ε eventually and −xk>−L−ε frequently. Negating each of the two inequalities reverses it, turning them into xk>L−ε eventually and xk<L+ε frequently, which is claim 2.

step 4.1L5L6∎

Remarks

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

A real sequence converges to L∈R iff lim inf⁡xk=lim sup⁡xk=L, and diverges to ±∞ iff both equal ±∞

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), with lim sup⁡ and lim inf⁡ as in Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾.

  1. For L∈R: (xk) converges to L (Limits and Cauchy sequences of reals) if and only if lim inf⁡kxk=lim sup⁡kxk=L.
  2. xk→+∞ (Divergence to +∞ and to −∞) if and only if lim inf⁡kxk=lim sup⁡kxk=+∞. Moreover lim inf⁡kxk=+∞ on its own already forces lim sup⁡kxk=+∞.
  3. xk→−∞ if and only if lim inf⁡kxk=lim sup⁡kxk=−∞, and lim sup⁡kxk=−∞ on its own already forces lim inf⁡kxk=−∞.

The three clauses combine into one statement about the extended line: for L∈R‾, the sequence (xk) converges to L in R‾ (Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞) if and only if

lim inf⁡kxk=lim sup⁡kxk=L.

Since lim inf⁡kxk≤lim sup⁡kxk always (lim inf⁡xk≤lim sup⁡xk for every real sequence), the single equation lim inf⁡kxk=lim sup⁡kxk is therefore equivalent to convergence in R‾, and the common value is the limit. A sequence that neither converges nor diverges to ±∞ is exactly one for which the inequality is strict.

Facts & Assumptions

Given: A sequence (xk) of reals, its tail ranges Tn={xk:k≥n}, the extended tail bounds sn=sup⁡Tn and in=inf⁡Tn, and the quantities lim sup⁡kxk=inf⁡{sn}, lim inf⁡kxk=sup⁡{in} (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L1]

All of sn, in, lim sup⁡kxk and lim inf⁡kxk exist in R‾ for every sequence; in is the greatest lower bound of Tn and lim inf⁡kxk the least upper bound of {in}, with the dual descriptions for sn and lim sup⁡kxk (The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence, Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R, Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

The order on R‾ is total, so the failure of a≤b is b<a; it restricts on R to the order of R; +∞ is the greatest element and −∞ the least; and every real is <+∞ and >−∞ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Epsilon characterisation, for a real L: L=lim sup⁡kxk exactly when for every real ε>0 one has xk<L+ε eventually and xk>L−ε frequently; and L=lim inf⁡kxk exactly when for every real ε>0 one has xk>L−ε eventually and xk<L+ε frequently (For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently).

[L4]

lim inf⁡kxk≤lim sup⁡kxk (lim inf⁡xk≤lim sup⁡xk for every real sequence).

[L5]

Reflection: lim sup⁡k(−xk)=−lim inf⁡kxk and lim inf⁡k(−xk)=−lim sup⁡kxk (lim sup⁡(−xk)=−lim inf⁡(xk), with the reflection of R‾ exchanging ±∞). Also xk→−∞ if and only if −xk→+∞: the condition xk<M for all k≥K is equivalent to −xk>−M for all k≥K by order reversal, and M runs over all reals exactly when −M does (Divergence to +∞ and to −∞); the order reversal used here is strict, and the form stated in Order is preserved by adding a constant and by adding inequalities is likewise strict, so nothing nonstrict is being borrowed from it.

[L6]

Convergence to a real L means: for every rational ε>0 there is K with ∣xk−L∣<ε for all k≥K; and the same relation is obtained by testing every real ε>0 instead, since below any positive real lies a positive rational (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The rationals embed densely in the reals).

[L7]

Divergence: xk→+∞ means that for every real M there is K with xk>M for all k≥K (Divergence to +∞ and to −∞).

[L8]

Eventually and frequently, and the fact that a property holding eventually holds frequently, since indices beyond any two given thresholds exist by totality of the order on N; likewise two properties each holding eventually hold together from the larger threshold on (Sequences of reals: bounded, eventually, frequently, tails, subsequences, ≤ is a linear order on N).

[L9]

Absolute value: for c>0, ∣a∣<c if and only if −c<a<c (Basic properties of the absolute value).

[L10]

Order arithmetic in R: 0<1, so t<t+1 for every real t, and no real is above every real; adding a constant preserves the order (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct
1.1

For the forward implication of claim 1, assume L∈R and that (xk) converges to L.

assume-hypL6
1.2

For the converse implication of claim 1, assume L∈R and lim inf⁡kxk=lim sup⁡kxk=L.

assume-hypL1
1.3

For the forward implication of claim 2, assume xk→+∞.

assume-hypL7
1.4

For the converse implication of claim 2, assume lim inf⁡kxk=+∞.

assume-hypL1
2.1

Under the assumption of step 1.1, let ε>0 be an arbitrary real. Testing convergence at ε gives K with ∣xk−L∣<ε, hence L−ε<xk<L+ε, for all k≥K. So xk<L+ε eventually and xk>L−ε eventually, and each of the two therefore also holds frequently. Both halves of each characterisation in [L3] are met, so lim sup⁡kxk=L and lim inf⁡kxk=L.

step 1.1L3L6L8L9
2.2

Under the assumption of step 1.2, let ε>0 be an arbitrary real. The forward halves of the two characterisations in [L3] give xk<L+ε for all k beyond some K1 and xk>L−ε for all k beyond some K2; beyond the larger of K1 and K2 both hold, so ∣xk−L∣<ε there. This holds for every real ε>0, in particular for every rational one, so (xk) converges to L.

step 1.2L3L6L8L9
2.3

Under the assumption of step 1.3, let M be an arbitrary real and take K with xk>M for all k≥K. Then M is a lower bound of TK, so M≤iK, and iK≤lim inf⁡kxk because lim inf⁡kxk is an upper bound of {in}; hence M≤lim inf⁡kxk. Since M was an arbitrary real, lim inf⁡kxk is not −∞, which lies below every real, and it is not a real t either, since M=t+1 would give t+1≤t. So lim inf⁡kxk=+∞.

step 1.3L1L2L7L10
2.4

Under the assumption of step 1.4, let M be an arbitrary real. Since sup⁡{in}=+∞ and M<+∞, the real M is not an upper bound of {in}, for otherwise the least upper bound would satisfy +∞≤M; by totality there is n with in>M. Every k≥n satisfies xk≥in>M, so xk>M eventually. As M was arbitrary, xk→+∞.

step 1.4L1L2L7
3.1

Steps 2.1 and 2.2 are the two implications of claim 1.

step 2.1step 2.2L3
3.2

For claim 2: if xk→+∞ then lim inf⁡kxk=+∞ by step 2.3, and then +∞=lim inf⁡kxk≤lim sup⁡kxk forces lim sup⁡kxk=+∞ since +∞ is the greatest element; conversely if lim inf⁡kxk=lim sup⁡kxk=+∞ then in particular lim inf⁡kxk=+∞ and step 2.4 gives xk→+∞. The same use of [L4] is the additional assertion that lim inf⁡kxk=+∞ alone forces lim sup⁡kxk=+∞.

step 2.3step 2.4L2L4
4.1

For claim 3, reflection gives xk→−∞ exactly when −xk→+∞, which by claim 2 holds exactly when lim inf⁡k(−xk)=lim sup⁡k(−xk)=+∞, that is −lim sup⁡kxk=−lim inf⁡kxk=+∞, that is lim sup⁡kxk=lim inf⁡kxk=−∞; and lim sup⁡kxk=−∞ alone forces lim inf⁡kxk≤−∞, hence lim inf⁡kxk=−∞, since −∞ is least. Claims 1, 2 and 3 together say that for L∈R‾ the sequence converges to L in R‾ exactly when lim inf⁡kxk=lim sup⁡kxk=L, since the three clauses of that definition are convergence to a real L, divergence to +∞ and divergence to −∞.

step 3.1step 3.2L2L4L5∎

Remarks

  • This is the theorem that makes lim sup⁡ and lim inf⁡ worth defining. They exist for every sequence, with no hypothesis, and their coincidence is exactly convergence in R‾. So a question about convergence becomes a question about two computable quantities, and a proof of convergence can be assembled from one-sided estimates without a candidate limit in hand.

  • The equation is between elements of R‾, and reading it in R would lose two thirds of the content. Clauses 2 and 3 are statements about divergence, and they are true statements about Divergence to +∞ and to −∞, not a redefinition of it: nothing above claims that a sequence diverging to +∞ has a limit in R, and the symbol +∞ occurring in them is the element of R‾ introduced in The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined.

  • A sequence with lim inf⁡<lim sup⁡ does neither. The alternating sequence is the standard witness, with the two values −1 and 1 ((−1)k has lim inf⁡=−1 and lim sup⁡=1, so it does not converge ↗); it is bounded, so it also does not diverge to ±∞, and the theorem says its failure to converge is exactly the gap between the two quantities.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The limit superior is itself a subsequential limit in R‾ and is the greatest one

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and write Λ:=lim sup⁡kxk∈R‾ (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾). Then, with the extended subsequential limit set SL⁡‾(x) of Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞:

  1. Λ∈SL⁡‾(x): there is a strictly increasing n:N→N such that (xnj) converges to Λ in R‾;
  2. L≤Λ for every L∈SL⁡‾(x).

So SL⁡‾(x) is nonempty and has a greatest element, and that element is lim sup⁡kxk. In particular every sequence of reals whatever has a subsequence that converges in R‾.

The extended set is the right home for this statement, and the real set is not. The finite subsequential limit set SL⁡(x) of Subsequential limit of a real sequence, and the subsequential limit set may be empty, and when it is not it may have a greatest element different from lim sup⁡kxk; both failures are exhibited by the dedicated counterexample on the companion page. What is true for SL⁡(x) follows: when Λ is a real number, claim 1 puts it in SL⁡(x), since the two sets agree on R (Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞), and claim 2 then makes it the greatest element there too.

Facts & Assumptions

Given: A sequence (xk) of reals, its tail ranges Tn={xk:k≥n}, the extended tail suprema sn=sup⁡Tn, and Λ:=lim sup⁡kxk=inf⁡{sn:n∈N} (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L2]

The order on R‾ is total, so the failure of a≤b is b<a; −∞ is least and +∞ greatest; every real is <+∞ and >−∞; and on R the order is that of R (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Epsilon characterisation for a real Λ: for every real η>0 one has xk<Λ+η eventually and xk>Λ−η frequently (For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently).

[L4]

Recursion theorem: for a set A, an element a∈A and a function f:A→A there is a unique g:N→A with g0=a and gj+1=f(gj) (The recursion theorem).

[L5]

Well-ordering principle: every nonempty subset of N has a least element (The well-ordering principle).

[L6]

Index maps: if nj<nj+1 for every j then n is strictly increasing, and then nj≥j for every j; the composite (xnj) is a subsequence (A strictly increasing index map satisfies nk≥k, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Convergence in R‾ and the extended subsequential limit set (Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞); convergence to a real, for which it suffices to produce a threshold for every real ε>0 (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences); divergence to ±∞ (Divergence to +∞ and to −∞); and ∣a−b∣<c if and only if b−c<a<b+c for c>0 (Basic properties of the absolute value).

[L9]

Limits preserve non-strict inequalities: if yj≤c for all large j and yj→y in R, then y≤c (Limits preserve non-strict inequalities).

[L10]

Archimedean facts: for every real M there is a natural p≥1 with M<p⋅1R, and for every real η>0 a natural m≥1 with 1/m<η; the canonical naturals satisfy 0≤n⋅1R and are increasing in n, and 0<a≤b gives 0<1/b≤1/a (Every complete ordered field is Archimedean, For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L11]

Strictly between any two reals lies a rational, hence a real (The rationals embed densely in the reals).

[L12]

The order on N is total and transitive, so any two indices have a common upper bound (Order on the natural numbers, ≤ is a linear order on N).

Proof

technique · constructive
1.1

The element Λ=lim sup⁡kxk exists in R‾, and exactly one of the following holds: Λ is a real number, Λ=+∞, or Λ=−∞.

givenL1L2
1.2

Suppose Λ=+∞. Since Λ is a lower bound of {sn}, every n has +∞≤sn and so sn=+∞. Consequently, for every n∈N and every real M there is k≥n with xk>M: otherwise M would be an upper bound of Tn and leastness would give sn≤M, contradicting M<+∞.

givenL1L2
1.3

Suppose Λ is real. Then for every n∈N and every real η>0 there is k≥n with ∣xk−Λ∣<η: by [L3] fix K with xk<Λ+η for all k≥K, let K′ be an index at least as large as both n and K, and use that xk>Λ−η frequently to obtain k≥K′ with xk>Λ−η; that k satisfies k≥K, hence also xk<Λ+η, and k≥n.

givenL3L7L12
1.4

Suppose Λ=−∞. Then xk→−∞ by [L8], and the identity map j↦j is strictly increasing, so the subsequence (xj) of (xk) converges to −∞ in R‾ and Λ∈SL⁡‾(x).

givenL6L7L8
1.5

Let L∈SL⁡‾(x) be arbitrary and fix a strictly increasing n:N→N such that (xnj) converges to L in R‾; then nj≥j for every j.

givenL6L7
2.1

In the case Λ=+∞, define f:N→N by letting f(n) be the least element of En:={ k∈N:k>n and xk>n⋅1R }, which is nonempty by step 1.2 applied with the index n+1 and the real M=n⋅1R, and let a be the least element of { k:xk>0 }, nonempty by step 1.2 with n=0 and M=0. Then f(n)>n and xf(n)>n⋅1R for every n.

step 1.2L5construct
2.2

In the case Λ real, define g:N→N by letting g(n) be the least element of Fn:={ k∈N:k>n and ∣xk−Λ∣<1/(n+1) }, which is nonempty by step 1.3 applied with the index n+1 and η=1/(n+1)>0, and let b be the least element of { k:∣xk−Λ∣<1 }, nonempty by step 1.3 with n=0 and η=1. Then g(n)>n and ∣xg(n)−Λ∣<1/(n+1) for every n.

step 1.3L5L10construct
2.3

If L=−∞ then L≤Λ, since −∞ is the least element of R‾.

step 1.5L2
2.4

If L=+∞, then for every real M there is J with xnj>M for all j≥J. Fix n∈N and a real M, and take j at least as large as both J and n; then nj≥j≥n, so xnj∈Tn and M<xnj≤sn. As M was an arbitrary real, sn is neither real nor −∞, so sn=+∞; as n was arbitrary, Λ=inf⁡{sn}=+∞ and L≤Λ.

step 1.5L1L2L7L12
2.5

If L is real, suppose for the sake of the comparison that Λ<L. By step 1.1 the element Λ is then real or −∞; choose a real c with Λ<c<L, taking a rational strictly between Λ and L in the first case and c:=L−1 in the second. Since Λ is the greatest lower bound of {sn} and Λ<c, the element c is not a lower bound, so there is n with sn<c, and then xk≤sn<c for every k≥n. For j≥n we have nj≥j≥n, hence xnj≤c, so L≤c by [L9], contradicting c<L. By totality L≤Λ.

step 1.5step 1.1L1L2L9L11
3.1

In the case Λ=+∞, the recursion theorem applied to N, the element a and the function f gives n:N→N with n0=a and nj+1=f(nj). Then nj<nj+1 for every j, so n is strictly increasing and nj≥j; and xnj+1>nj⋅1R≥j⋅1R for every j.

step 2.1L4L6L10
3.2

In the case Λ real, the recursion theorem applied to N, the element b and the function g gives n:N→N with n0=b and nj+1=g(nj). Then n is strictly increasing with nj≥j, and ∣xnj+1−Λ∣<1/(nj+1)≤1/(j+1) for every j.

step 2.2L4L6L10
4.1

In the case Λ=+∞, the subsequence (xnj) diverges to +∞: given a real M, take a natural p≥1 with M<p⋅1R; every j≥p+1 satisfies j−1≥p, so step 3.1 applied at j−1 gives xnj>(j−1)⋅1R≥p⋅1R>M. Hence (xnj) converges to +∞=Λ in R‾ and Λ∈SL⁡‾(x).

step 3.1L7L10L12
4.2

In the case Λ real, the subsequence (xnj) converges to Λ: given a real ε>0, take a natural m≥1 with 1/m<ε; every j≥m satisfies j≥1, so step 3.2 applied at j−1 gives ∣xnj−Λ∣<1/j≤1/m<ε. Producing such a threshold for every real ε>0 establishes convergence, so (xnj) converges to Λ in R‾ and Λ∈SL⁡‾(x).

step 3.2L7L10
5.1

The three cases of step 1.1 are exhaustive, and each produces a subsequence converging to Λ in R‾: step 4.1 when Λ=+∞, step 4.2 when Λ is real, and step 1.4 when Λ=−∞. So Λ∈SL⁡‾(x), which is claim 1.

step 4.1step 4.2step 1.4L7
6.1

Steps 2.3, 2.4 and 2.5 cover the three possibilities for an arbitrary L∈SL⁡‾(x) and give L≤Λ in each, which is claim 2. With claim 1 this makes SL⁡‾(x) nonempty with greatest element Λ=lim sup⁡kxk.

step 5.1step 2.3step 2.4step 2.5L2discharge-construct∎

Remarks

  • The construction uses no choice. Both index maps are built by taking a least element (The well-ordering principle) of an explicitly described nonempty set of naturals, so the functions f and g are defined outright and The recursion theorem then produces the index map. This is the same device as in Every real sequence has a monotone subsequence (the peak / rising-sun lemma), and for the same reason: a subsequence selected by repeated arbitrary choices would need a choice principle, and none is needed here.

  • Why the recursion threshold is indexed by the previous index rather than by the step number. The recursion theorem produces a function of one variable, so the state carried from one step to the next is the index nj alone. Demanding xnj+1>nj rather than xnj+1>j keeps that single-variable form, and nj≥j (A strictly increasing index map satisfies nk≥k) then upgrades the bound to the one actually wanted. The same trick fixes the accuracy in the finite case at 1/(nj+1)≤1/(j+1).

  • Claim 2 is where the lim sup⁡ earns the word "greatest". A subsequence cannot do better than the tail suprema allow: past any index n, every term of the sequence, and so every term of any subsequence, is at most sn, and Λ is the infimum of those. That is the entire content of step 2.5, and the strictness of the inequality Λ<c is what gives the contradiction, since a limit inherits only the non-strict inequality (Limits preserve non-strict inequalities).

  • Both failures of the real version really occur, and A sequence with lim sup⁡=+∞: the greatest subsequential limit exists only in R‾ ↗ on the companion page is the witness: there SL⁡(x) is nonempty with greatest element 0 while lim sup⁡kxk=+∞.

  • The dual statement is The limit inferior is the least subsequential limit in R‾, obtained from this theorem by reflection rather than by repeating the construction.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The limit inferior is the least subsequential limit in R‾

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences). Then lim inf⁡kxk∈SL⁡‾(x) and lim inf⁡kxk≤L for every L∈SL⁡‾(x) (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, Convergence in R‾ and the extended subsequential limit set: L∈R‾ is an extended subsequential limit when some subsequence converges to L, or diverges to L=±∞).

So the extended subsequential limit set of any real sequence has a least element as well as a greatest one, and the two are lim inf⁡kxk and lim sup⁡kxk respectively (The limit superior is itself a subsequential limit in R‾ and is the greatest one). Every extended subsequential limit lies between them.

Facts & Assumptions

Given: A sequence (xk) of reals, and its reflection yk:=−xk.

[L1]

Reflection on R‾: a↦−a satisfies −(−a)=a and a≤b if and only if −b≤−a (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

[L2]
[L3]

For every real sequence the extended subsequential limit set is nonempty and has greatest element the limit superior (The limit superior is itself a subsequential limit in R‾ and is the greatest one).

[L5]

Scalar multiples of convergent sequences: zj→z in R implies czj→cz (Algebra of limits: sums, scalar multiples, products and quotients).

[L6]

Divergence to ±∞, and order reversal: zj>M is equivalent to −zj<−M, and M runs over all reals exactly when −M does (Divergence to +∞ and to −∞, Order is preserved by adding a constant and by adding inequalities).

Proof

technique · direct
1.1

Put yk:=−xk, a sequence of reals; then −yk=xk for every k, by the involution property of the reflection.

givenL1L4
1.2

Let L∈R‾ and let n:N→N be strictly increasing with (xnj) converging to L in R‾.

givenL4
1.3

By [L3] applied to the sequence (yk), the set SL⁡‾(y) is nonempty and has greatest element N0:=lim sup⁡kyk, and N0=−lim inf⁡kxk by [L2].

givenL2L3L7
2.1

The reflected subsequence (ynj)=(−xnj) converges to −L in R‾. If L is real this is the scalar rule with c=−1. If L=+∞ then for every real M there is J with xnj>M for all j≥J, hence ynj<−M for all such j; since −M runs over all reals as M does, ynj→−∞=−L. If L=−∞ the same argument with the inequalities exchanged gives ynj→+∞=−L.

step 1.2L1L4L5L6
3.1

Hence L∈SL⁡‾(x) implies −L∈SL⁡‾(y), the same index map serving. Applying that implication to the sequence (yk), whose reflection is (xk), gives conversely that N∈SL⁡‾(y) implies −N∈SL⁡‾(x). So SL⁡‾(x)={ −N:N∈SL⁡‾(y) }.

step 2.1step 1.1L1L4
4.1

Therefore −N0∈SL⁡‾(x), and −N0=−(−lim inf⁡kxk)=lim inf⁡kxk; and for any L∈SL⁡‾(x) the element −L lies in SL⁡‾(y), so −L≤N0 by maximality, whence lim inf⁡kxk=−N0≤L by order reversal. Thus lim inf⁡kxk is the least element of SL⁡‾(x).

step 3.1step 1.3L1L2∎

Remarks

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

If each yj is a subsequential limit of (xk) and yj→y∈R, then y is a subsequential limit of (xk)

Statement

Let (xk) be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and let (yj) be a sequence of reals with

Then y∈SL⁡(x).

In words: the subsequential limit set of a real sequence contains the limit of every convergent sequence of its own elements. When the topology of R arrives, that property is what it calls sequential closedness; that sequential closedness is in turn equivalent to closedness for subsets of R is a theorem there and not a matter of naming, and the half of that equivalence running from sequential closedness to closedness spends the axiom of countable choice. Here the property is stated and proved purely in terms of sequences, with no choice principle and no topological notion used or needed.

Facts & Assumptions

Given: A sequence (xk) of reals; a sequence (yj) of reals with yj∈SL⁡(x) for every j; and a real y with yj→y.

[L1]

Subsequential limits and convergence: yj∈SL⁡(x) means that some strictly increasing m:N→N has xmi→yj; convergence of a sequence of reals is the rational-ε condition of Limits and Cauchy sequences of reals, and to establish convergence it suffices to produce a threshold for every real ε>0, by the remark of Sequences of reals: bounded, eventually, frequently, tails, subsequences (Subsequential limit of a real sequence, and the subsequential limit set, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L2]

Index maps: a strictly increasing m satisfies mi≥i, and an index map with nj<nj+1 for every j is strictly increasing (A strictly increasing index map satisfies nk≥k).

[L3]

Well-ordering principle: every nonempty subset of N has a least element (The well-ordering principle).

[L4]

Recursion theorem: for a set A, an element a∈A and f:A→A there is a unique g:N→A with g0=a and gj+1=f(gj) (The recursion theorem).

[L5]

Absolute value and the triangle inequality: ∣a+b∣≤∣a∣+∣b∣, and ∣a∣<c if and only if −c<a<c for c>0 (The triangle inequality, Basic properties of the absolute value).

[L6]

Canonical naturals and reciprocals: for a natural q≥1 the element q⋅1R is positive and invertible, (2(n+1))⋅1R=2((n+1)⋅1R), and 0<a≤b gives 0<1/b≤1/a; moreover for every real η>0 there is a natural m≥1 with 1/m<η (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean).

[L7]

The order on N is total, so any two indices have a common upper bound (Order on the natural numbers, ≤ is a linear order on N).

[L8]

Below every positive real lies a positive rational, which is how a convergence hypothesis stated for rational ε is instantiated at a real threshold (The rationals embed densely in the reals).

Proof

technique · constructive
1.1

For n∈N put qn:=2(n+1), a natural number ≥1. Then qn⋅1R=2((n+1)⋅1R)>0 is invertible, 1/qn>0, and 1/qn+1/qn=2/qn=1/(n+1).

givenL6algebra
1.2

By hypothesis (yj) converges to y and every yj lies in SL⁡(x), so for each j there is a strictly increasing m with xmi→yj.

givenL1
2.1

For every n∈N there is k>n with ∣xk−y∣<1/(n+1). Indeed, take a rational ε1 with 0<ε1<1/qn and instantiate the convergence yj→y at ε1 to obtain an index J with ∣yJ−y∣<ε1<1/qn. Since yJ∈SL⁡(x), fix a strictly increasing m with xmi→yJ, take a rational ε2 with 0<ε2<1/qn and an index I with ∣xmi−yJ∣<ε2 for all i≥I, and let i be an index at least as large as both I and n+1. Then k:=mi satisfies k=mi≥i≥n+1>n and, by the triangle inequality applied to xk−y=(xk−yJ)+(yJ−y), ∣xk−y∣≤∣xk−yJ∣+∣yJ−y∣<1/qn+1/qn=1/(n+1).

step 1.1step 1.2L1L2L5L7L8
3.1

Define f:N→N by letting f(n) be the least element of the set Gn:={ k∈N:k>n and ∣xk−y∣<1/(n+1) }, which is nonempty by step 2.1. Then f(n)>n and ∣xf(n)−y∣<1/(n+1) for every n.

step 2.1L3construct
4.1

The recursion theorem applied to N, the element f(0) and the function f gives n:N→N with n0=f(0) and nj+1=f(nj). Then nj<nj+1 for every j, so n is strictly increasing and nj≥j; moreover ∣xnj+1−y∣<1/(nj+1)≤1/(j+1) for every j, using that nj+1≥j+1.

step 3.1L2L4L6
5.1

The subsequence (xnj) converges to y: given a real ε>0, take a natural m≥1 with 1/m<ε; every j≥m satisfies j≥1, so step 4.1 applied at j−1 gives ∣xnj−y∣<1/j≤1/m<ε. Producing such a threshold for every real ε>0 establishes convergence, and n is strictly increasing, so y∈SL⁡(x).

step 4.1L1L2L6discharge-construct∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

If xk≤yk eventually then lim sup⁡xk≤lim sup⁡yk and lim inf⁡xk≤lim inf⁡yk

Statement

Let (xk) and (yk) be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with xk≤yk eventually, that is for all k from some index on. Then

lim sup⁡kxk  ≤  lim sup⁡kykandlim inf⁡kxk  ≤  lim inf⁡kyk

in R‾ (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined). No boundedness or convergence hypothesis is placed on either sequence.

Facts & Assumptions

Given: Sequences (xk) and (yk) of reals and an index K∈N with xk≤yk for every k≥K; the tail ranges Tn(x)={xk:k≥n} and Tn(y), and the extended tail bounds sn(x)=sup⁡Tn(x), in(x)=inf⁡Tn(x) and likewise for y (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L1]

All tail bounds and both of lim sup⁡, lim inf⁡ exist in R‾; sn is the least upper bound of the tail range and in its greatest lower bound; lim sup⁡kyk is the greatest lower bound of {sn(y)} and lim inf⁡kyk the least upper bound of {in(y)}; and sm≤sn, in≤im whenever n≤m (The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence, Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R, Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L3]

A property holds eventually when it holds for all indices from some index on (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

The order on N is total, so every n satisfies n≥K or n<K, and in the latter case n≤K (Order on the natural numbers, ≤ is a linear order on N).

Proof

technique · direct
1.1

By hypothesis fix K∈N with xk≤yk for every k≥K.

givenL3
2.1

Let n≥K. Every k≥n satisfies k≥K, so xk≤yk≤sn(y), and therefore sn(y) is an upper bound of Tn(x), whence sn(x)≤sn(y) by leastness. Dually in(x)≤xk≤yk for every k≥n, so in(x) is a lower bound of Tn(y) and in(x)≤in(y) by greatest-lower-boundedness.

step 1.1L1L2L4
3.1

For every n∈N one has lim sup⁡kxk≤sn(y). If n≥K this is lim sup⁡kxk≤sn(x)≤sn(y), the first inequality because lim sup⁡kxk is a lower bound of {sm(x)}. If n<K then n≤K, so sK(y)≤sn(y), and lim sup⁡kxk≤sK(x)≤sK(y)≤sn(y).

step 2.1L1L2L4
3.2

For every n∈N one has in(x)≤lim inf⁡kyk. If n≥K this is in(x)≤in(y)≤lim inf⁡kyk, the second inequality because lim inf⁡kyk is an upper bound of {im(y)}. If n<K then n≤K, so in(x)≤iK(x)≤iK(y)≤lim inf⁡kyk.

step 2.1L1L2L4
4.1

By step 3.1 the element lim sup⁡kxk is a lower bound of {sn(y):n∈N}, whose greatest lower bound is lim sup⁡kyk, so lim sup⁡kxk≤lim sup⁡kyk. By step 3.2 the element lim inf⁡kyk is an upper bound of {in(x):n∈N}, whose least upper bound is lim inf⁡kxk, so lim inf⁡kxk≤lim inf⁡kyk.

step 3.1step 3.2L1∎

Remarks

  • "Eventually" is enough, and the proof shows why. Only tails with n≥K are compared directly; the finitely many earlier tail bounds are absorbed by monotonicity of the tail bounds (The tail suprema of any real sequence are nonincreasing in R‾, so the limit superior exists for every sequence), which lets sK(y) stand in for every earlier sn(y). No appeal to Convergence depends only on the tail is needed, since neither quantity is defined as a limit.

  • The comparison does not become strict. From xk<yk for every k one gets only lim sup⁡kxk≤lim sup⁡kyk; the sequences xk=0 and yk=1/(k+1) have equal limits and hence equal limit superiors. This is the same phenomenon as for limits (Limits preserve non-strict inequalities).

  • Both conclusions have the same direction. It is the inner operation that differs between lim sup⁡ and lim inf⁡, and both a supremum and an infimum are monotone in the set, so a pointwise inequality pushes both quantities the same way. What fails to be monotone is the gap between them: nothing here compares lim sup⁡kxk with lim inf⁡kyk.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

lim sup⁡(xk+yk)≤lim sup⁡xk+lim sup⁡yk whenever the right-hand side is defined in R‾, and dually for lim inf⁡

Statement

Let (xk) and (yk) be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and write Λ:=lim sup⁡kxk, M:=lim sup⁡kyk (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

  1. If the sum Λ+M is defined in R‾ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined), that is if {Λ,M}≠{+∞,−∞}, then lim sup⁡k(xk+yk)  ≤  Λ+M.
  2. Dually, writing λ:=lim inf⁡kxk and μ:=lim inf⁡kyk, if λ+μ is defined in R‾ then lim inf⁡k(xk+yk)  ≥  λ+μ.

The hypothesis is exactly the one The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined forces, and it cannot be dropped. When one of Λ, M is +∞ and the other −∞ the right-hand side is not an element of R‾ at all, so there is nothing to compare. The inequality is genuinely an inequality: equality can fail, and does, for an alternating pair of sequences; the failure of additivity is recorded as a false statement among this page's examples, and the witness is a named counterexample on the companion page.

Facts & Assumptions

Given: Sequences (xk) and (yk) of reals, their termwise sum (xk+yk), and Λ:=lim sup⁡kxk, M:=lim sup⁡kyk, assumed to have a sum defined in R‾.

[L2]

The order on R‾ is total and transitive, +∞ is its greatest element and −∞ its least, and it restricts on R to the order of R (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Partial addition on R‾: a sum is undefined only for the pairs (+∞,−∞) and (−∞,+∞); a sum with one summand +∞ and the other ≠−∞ is +∞; a sum with one summand −∞ and the other ≠+∞ is −∞; and −(a+b)=(−a)+(−b), each side defined exactly when the other is (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

[L4]

Epsilon characterisation for a real limit superior: Λ=lim sup⁡kxk real implies that for every real ε>0 one has xk<Λ+ε eventually (For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently).

[L6]

Reflection: lim sup⁡k(−zk)=−lim inf⁡kzk and lim inf⁡k(−zk)=−lim sup⁡kzk (lim sup⁡(−xk)=−lim inf⁡(xk), with the reflection of R‾ exchanging ±∞).

[L7]

Order arithmetic in R: Order is preserved by adding a constant and by adding inequalities states the strict forms, that inequalities may be translated and added, so a<a′ and b<b′ give a+b<a′+b′; adjoining the case of equality, in which both sides move by the same amount, gives the nonstrict forms used below. In particular a≤b if and only if −b≤−a: translation by −a−b turns a<b into −b<−a and back, while a=b holds exactly when −a=−b.

[L8]

Reciprocal Archimedean property and canonical naturals: for every real δ>0 there is a natural m≥1 with 1/m<δ; for a natural m≥1 the element 2m is a natural ≥1 with (2m)⋅1R=2(m⋅1R)>0, so 1/(2m)>0 and 1/(2m)+1/(2m)=1/m (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order, Canonical naturals are positive and strictly increasing).

[L9]

Two properties each holding eventually hold together from the larger of the two thresholds on (Sequences of reals: bounded, eventually, frequently, tails, subsequences, ≤ is a linear order on N, Order on the natural numbers).

Proof

technique · direct
1.1

Since Λ+M is defined, exactly one of the following three situations holds: at least one of Λ, M equals +∞, and then the other is ≠−∞; both are real; or neither equals +∞ and at least one equals −∞. Both the hypothesis and the conclusion of claim 1 are unchanged by exchanging the two sequences, so in the third situation it may be assumed that Λ=−∞.

givenL2L3
2.1

In the first situation Λ+M=+∞ by the addition table, and every element of R‾ is ≤+∞, so lim sup⁡k(xk+yk)≤Λ+M.

step 1.1L2L3
2.2

In the second situation let δ>0 be an arbitrary real, take a natural m≥1 with 1/m<δ and put ε:=1/(2m)>0, so that ε+ε=1/m<δ. By [L4] there are thresholds beyond which xk<Λ+ε and beyond which yk<M+ε; beyond the larger of them both hold, so adding the two inequalities gives xk+yk<Λ+M+ε+ε for all k≥N, where N is that larger threshold. Hence Λ+M+ε+ε is an upper bound of the N-th tail range of (xk+yk), so the N-th tail supremum is ≤Λ+M+ε+ε, and therefore lim sup⁡k(xk+yk)≤Λ+M+ε+ε<Λ+M+δ.

step 1.1L1L2L4L7L8L9
2.3

In the third situation, with Λ=−∞, first note that there is a real B with yk<B eventually: if M is real, [L4] with ε=1 gives yk<M+1 eventually, so B:=M+1 serves; and if M=−∞ then yk→−∞ by [L5], so yk<0 eventually and B:=0 serves. Also Λ=−∞ gives xk→−∞ by [L5]. Now let c be an arbitrary real: since c−B is real, xk<c−B eventually, and beyond the larger threshold both that and yk<B hold, so xk+yk<(c−B)+B=c there. As c was arbitrary, xk+yk→−∞, hence lim sup⁡k(xk+yk)=−∞=Λ+M by [L5] and the addition table.

step 1.1L3L4L5L7L9
3.1

In the second situation the conclusion follows from step 2.2: taking δ=1 shows lim sup⁡k(xk+yk)≤Λ+M+1, a real number, so the left-hand side is not +∞; if it is −∞ then it is ≤Λ+M because −∞ is least; and if it is a real S with S>Λ+M, then δ0:=S−(Λ+M)>0 and step 2.2 applied with δ=δ0 gives S<Λ+M+δ0=S, which is impossible. So lim sup⁡k(xk+yk)≤Λ+M by totality.

step 2.2L2L7
4.1

Claim 1 now holds in all three situations, by steps 2.1, 3.1 and 2.3.

step 2.1step 3.1step 2.3step 1.1
5.1

For claim 2, suppose λ+μ is defined. By [L6] the reflected sequences have lim sup⁡k(−xk)=−λ and lim sup⁡k(−yk)=−μ, and (−λ)+(−μ)=−(λ+μ) is defined exactly when λ+μ is, by [L3]. Claim 1 applied to (−xk) and (−yk), whose termwise sum is (−(xk+yk)), therefore gives −lim inf⁡k(xk+yk)=lim sup⁡k(−(xk+yk))≤(−λ)+(−μ)=−(λ+μ); reflecting this inequality reverses it into lim inf⁡k(xk+yk)≥λ+μ.

step 4.1L3L6L7∎

Remarks

  • The three situations are not decoration. The middle one is the analytic content and the outer two are genuinely different arguments: the first is vacuous because +∞ bounds everything, and the third is a statement about divergence to −∞ that has to be proved, since a sum of two sequences each running off to −∞, or one running off with the other merely bounded above, is not covered by any algebra of limits (Divergence to +∞ and to −∞ forbids that).

  • Why the real supremum of a sumset is not used. The natural one-line route, sn(x+y)≤sn(x)+sn(y) followed by a passage to the infimum, needs the first inequality in R‾ and then still needs an ε argument to compare inf⁡n(sn(x)+sn(y)) with Λ+M. The identity sup⁡(S+T)=sup⁡S+sup⁡T of Supremum of a sumset: sup⁡(S+T)=sup⁡S+sup⁡T does not apply, since it requires both sets to be nonempty subsets of R bounded above, and a tail range of an unbounded sequence is not. The ε argument is therefore made directly, once.

  • Both halves of the ε split are reciprocals of natural numbers, not halvings in R. Choosing m with 1/m<δ and then working with 1/(2m) keeps every quantity a reciprocal of a canonical natural, so the only field facts used are that positives are invertible and that inequalities add.

  • Equality is the exception. Without a hypothesis on one of the two sequences the gap can be as large as the whole oscillation, as xk=(−1)k, yk=(−1)k+1 give lim sup⁡(xk+yk)=0<2=lim sup⁡xk+lim sup⁡yk ↗ shows. It is standard, and neither needed nor proved on this page, that the inequality becomes an equality as soon as one of the two sequences converges to a real limit.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

For bounded nonnegative sequences, lim sup⁡(xkyk)≤(lim sup⁡xk)(lim sup⁡yk)

Statement

Let (xk) and (yk) be bounded sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with xk≥0 and yk≥0 for every k∈N. Then lim sup⁡kxk, lim sup⁡kyk and lim sup⁡k(xkyk) are real numbers, all ≥0, and

lim sup⁡k(xkyk)  ≤  (lim sup⁡kxk)(lim sup⁡kyk).

Both hypotheses are doing work. Boundedness makes all three quantities real, so that the product on the right is a product in the field R and no extended multiplication is involved; without it the right-hand side could be an undefined product 0⋅(+∞) (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined). Nonnegativity is what lets two upper estimates be multiplied: for sequences of mixed sign the inequality is false in the stated form, since a product of two negative numbers is positive and the estimate would point the wrong way. Strictness is possible, and a witness is recorded on the companion page.

Facts & Assumptions

Given: Bounded sequences (xk), (yk) of reals with xk≥0 and yk≥0 for every k; their termwise product (xkyk); and Λ:=lim sup⁡kxk, M:=lim sup⁡kyk, P:=lim sup⁡k(xkyk) (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

[L2]

The order on R‾ is total and transitive, restricts on R to the order of R, and has +∞ greatest and −∞ least; a member of R‾ lying between two reals is itself real (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Epsilon characterisation for a real limit superior: for every real ε>0 one has zk<lim sup⁡kzk+ε eventually (For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently).

[L4]

Boundedness of a sequence of reals: there is a real B with ∣zk∣≤B for every k; and z≤∣z∣ always (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Lower bound, bounded below, bounded set, Basic properties of the absolute value).

[L5]

Products of inequalities: 0≤a≤b and 0≤c≤d give ac≤bd, which Multiplying inequalities of positives states in exactly this nonstrict form; and multiplication by a positive element preserves the order, Sign rules for products and monotonicity of multiplication stating the strict form a<b  ⟺  ac<bc and the nonstrict form following by adjoining the case a=b, where the two products are equal.

[L6]

Order arithmetic in R: inequalities may be added and translated, and the order is total, so exactly one of a<b, a=b, b<a holds (Order is preserved by adding a constant and by adding inequalities).

[L7]

Reciprocal Archimedean property: for every real η>0 there is a natural m≥1 with 1/m<η; and 0<a≤b gives 0<1/b≤1/a (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L8]

Two properties each holding eventually hold together from the larger of the two thresholds on, the order on N being total (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Order on the natural numbers, ≤ is a linear order on N).

Proof

technique · direct
1.1

Both sequences are bounded, so there are reals bounding ∣xk∣ and ∣yk∣; let B be the larger of the two, so that ∣xk∣≤B and ∣yk∣≤B for every k, and B≥∣x0∣≥0. With xk≥0 and yk≥0 this gives 0≤xk≤B and 0≤yk≤B for every k, hence 0≤xkyk≤B⋅B by [L5].

givenL4L5L6
2.1

Each of Λ, M, P is a real number ≥0. Indeed, for every n the real B is an upper bound of the n-th tail range of (xk), so sn≤B and hence Λ≤s0≤B; and sn≥xn≥0 for every n, so 0 is a lower bound of {sn} and 0≤Λ. Being between the reals 0 and B, the element Λ is real. The same argument gives 0≤M≤B, and, using the bound B⋅B from step 1.1, 0≤P≤B⋅B.

step 1.1L1L2
3.1

Let δ>0 be an arbitrary real and put C:=Λ+M+1, a real with C≥1>0. Take a natural m1≥1 with 1/m1<1 and a natural m2≥1 with 1/m2<δ/C, let m be the larger of m1 and m2, and set ε:=1/m, so that 0<ε<1 and εC<δ. By [L3] there are thresholds beyond which xk<Λ+ε and beyond which yk<M+ε; let N be the larger. For k≥N we have 0≤xk≤Λ+ε and 0≤yk≤M+ε, so xkyk≤(Λ+ε)(M+ε)=ΛM+ε(Λ+M+ε)≤ΛM+εC<ΛM+δ, the middle step because Λ+M+ε≤C and ε>0. Hence ΛM+εC is an upper bound of the N-th tail range of (xkyk), so P≤ΛM+εC<ΛM+δ.

step 2.1L1L3L5L6L7L8algebra
4.1

Suppose P>ΛM. Both are real by step 2.1, so δ0:=P−ΛM>0, and step 3.1 applied with δ=δ0 gives P<ΛM+δ0=P, which is impossible. By totality P≤ΛM, which is the asserted inequality.

step 3.1step 2.1L2L6∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

n1/n→1

Statement

For a natural number n≥1 write ι(n):=n⋅1R for the canonical natural of R (Canonical naturals are positive and strictly increasing) and n1/n:=ι(n)1/n, n1/2:=ι(n)1/2 for its roots (Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Rational powers ar of a positive base). Then:

  1. 1  ≤  n1/n  ≤  1+2n1/2 for every natural n≥1;
  2. the sequence rk:=(k+1)1/(k+1), k∈N, converges to 1 (Limits and Cauchy sequences of reals).

The index range is not cosmetic. The expression n1/n is defined only for n≥1, since 1/n is not a rational number when n=0 (Rational powers ar of a positive base). Sequences in this library are functions on N and N contains 0 (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so the statement of convergence is made about the shifted family rk=(k+1)1/(k+1), which is the classical family n1/n, n≥1, reindexed by n=k+1. Claim 1 is stated over the natural range n≥1 where the expression means something.

Facts & Assumptions

Given: For a natural m≥1 the canonical natural ι(m):=m⋅1R, extended by ι(0):=0; this extension keeps the additivity ι(m+m′)=ι(m)+ι(m′) of Canonical naturals are positive and strictly increasing, which for m or m′ equal to 0 reads ι(m)=ι(m)+0.

[L1]

Roots: for real a≥0 and natural n≥1 there is a unique real s≥0 with sn=a, written a1/n; it is >0 when a>0, and a1/1=a (Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Integer powers am).

[L2]

Rational powers and monotonicity: a1/n is the rational power ar at r=1/n, and for rational t>0 one has at>1 whenever a>1; also (1/a)1/2=1/a1/2 for a>0 (Rational powers ar of a positive base, Monotonicity of r↦ar and of a↦ar, Laws of rational exponents).

[L3]

AM-GM: for a natural n≥1 and reals a0,…,an−1≥0, the geometric mean (∏j<naj)1/n is ≤ the arithmetic mean 1ι(n)∑j<naj (The arithmetic mean, geometric mean inequality).

[L4]

Finite sums and products: the empty sum is 0 and the empty product 1; sums and products split at any intermediate index; and ∑j<mλ=ι(m)λ for a constant λ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]
[L6]

Canonical naturals: ι(m)>0 and ι(m) is invertible for m≥1, ι is strictly increasing, and ι(2)=2>1; the Archimedean property gives, for every real x, a natural p≥1 with x<ι(p) (Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean).

[L7]

Order and reciprocals: 0<a<b gives 0<1/b<1/a; multiplying an inequality by a positive element preserves it; and inequalities may be added and translated (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities).

[L8]

Squares: for a,b≥0 one has a<b if and only if a⋅a<b⋅b (Monotonicity of x↦xn and of n↦an, Integer powers am).

[L9]

Squeeze theorem, and the fact that a constant sequence converges to its value; to establish convergence it suffices to produce a threshold for every real ε>0 (The squeeze theorem, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L10]

The order on N is total and ι respects it (Order on the natural numbers, ≤ is a linear order on N).

Proof

technique · direct
1.1

For a natural n≥1 the element ι(n) is positive and invertible, so ι(n)1/n and ι(n)1/2 exist and are positive.

givenL1L6
1.2

For every natural m one has ∏j<m1=1: the empty product is 1, and if ∏j<m1=1 then ∏j<m+11=(∏j<m1)⋅1=1, so this follows by induction on m.

givenL4L5
2.1

For n=1 one has ι(1)=1 and 11/1=1; for n≥2 one has ι(n)≥ι(2)=2>1 and 1/n is a positive rational, so ι(n)1/n>1. In either case n1/n≥1.

step 1.1L1L2L6L10
2.2

Let n≥2 and put u:=ι(n)1/2, so that u>0 and u⋅u=ι(n). Apply [L3] to the list of n nonnegative reals given by a0=a1=u and aj=1 for 2≤j<n, the latter range being empty when n=2. Splitting at index 2 gives ∏j<naj=(∏j<2aj)(∏j<n−2a2+j)=(u⋅u)⋅1=ι(n) by step 1.2, so the geometric mean is ι(n)1/n; and ∑j<naj=(∑j<2aj)+(∑j<n−21)=(u+u)+ι(n−2)=(u+u)+ι(n)−2, using additivity of ι and ι(2)=2, so the arithmetic mean is A=((u+u)+ι(n)−2)/ι(n)=1+((u+u)−2)/ι(n). Since (u+u)−2<u+u and ι(n)>0, and (u+u)/ι(n)=(u+u)/(u⋅u)=2/u, this gives ι(n)1/n≤A≤1+2/u=1+2/n1/2.

step 1.1step 1.2L1L3L4L6L7algebra
2.3

For n=1 the same bound holds trivially: 11/1=1≤1+2=1+2/11/2.

step 1.1L1L6L7
2.4

The sequence bk:=1+2/(k+1)1/2 converges to 1. Given a real ε>0, put t:=2/ε>0 and take a natural p≥1 with t⋅t<ι(p). For k≥p we have k+1>p, hence ι(k+1)>ι(p)>t⋅t, and since (ι(k+1)1/2)(ι(k+1)1/2)=ι(k+1) with both factors ≥0, this forces t<ι(k+1)1/2. Therefore 0<2/ι(k+1)1/2<2/t=ε, that is ∣bk−1∣<ε.

step 1.1L1L6L7L8L9L10algebra
3.1

Claim 1 is the combination of steps 2.1, 2.2 and 2.3, the two upper bounds covering n≥2 and n=1 respectively.

step 2.1step 2.2step 2.3
4.1

For every k∈N the natural k+1 is ≥1, so claim 1 gives 1≤rk≤bk. The constant sequence 1 converges to 1 and (bk) converges to 1 by step 2.4, so the squeeze theorem gives rk→1, which is claim 2.

step 3.1step 2.4L9∎

Remarks

  • Where the n comes from. AM-GM is applied to a list whose product is n but whose entries are as close to 1 as possible: two copies of n1/2 and n−2 copies of 1. The arithmetic mean is then 1+(2n1/2−2)/n, which tends to 1 at the rate 2/n1/2. Splitting n as n1/2⋅n1/2 rather than as n⋅1 is the whole trick: the list n,1,…,1 gives only n1/n≤2−1/n, which does not converge to 1.

  • The lower bound is not decoration. Without n1/n≥1 the squeeze has nothing below it, and the upper bound alone would leave open a limit smaller than 1. It comes from monotonicity of rational powers in the base (Monotonicity of r↦ar and of a↦ar) and holds with equality only at n=1.

  • No logarithm and no exponential is used. The usual quick proof writes n1/n=e(log⁡n)/n and appeals to (log⁡n)/n→0; neither function exists in this library yet, and the AM-GM route needs nothing beyond roots and finite sums.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

For every a>0, a1/n→1

Statement

Let a∈R with a>0, write ι(n):=n⋅1R for the canonical natural (Canonical naturals are positive and strictly increasing) and a1/n for the n-th root (Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Rational powers ar of a positive base), defined for naturals n≥1. Then:

  1. for every real b≥1 and every natural n≥1, 1  ≤  b1/n  ≤  1+b−1ι(n);
  2. the sequence ck:=a1/(k+1), k∈N, converges to 1 (Limits and Cauchy sequences of reals).

Index range. As for the previous lemma on this page, a1/n requires n≥1, so the sequence indexed by N (Sequences of reals: bounded, eventually, frequently, tails, subsequences) is the shifted family a1/(k+1); it is the classical family a1/n, n≥1, reindexed by n=k+1.

Facts & Assumptions

Given: A real a>0; the canonical naturals ι(n)=n⋅1R for n≥1; and the sequence ck:=a1/(k+1).

[L1]

Roots: for real x≥0 and natural n≥1 there is a unique real s≥0 with sn=x, written x1/n; it is >0 when x>0, and 11/n=1 by uniqueness (Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Integer powers am).

[L2]

Rational powers: x1/n is the rational power at exponent 1/n; for rational t>0, x>1 implies xt>1; and (xy)1/n=x1/ny1/n for x,y>0 (Rational powers ar of a positive base, Monotonicity of r↦ar and of a↦ar, Laws of rational exponents).

[L3]

Bernoulli's inequality: (1+x)n≥1+ι(n)x for x≥−1 and n∈N (Bernoulli's inequality (1+x)n≥1+nx).

[L4]

Canonical naturals: ι(n)>0 and invertible for n≥1, and ι is strictly increasing (Canonical naturals are positive and strictly increasing, Order on the natural numbers, ≤ is a linear order on N).

[L5]

Reciprocal Archimedean property: for every real η>0 there is a natural p≥1 with 1/p<η; and 0<x<y gives 0<1/y<1/x (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L6]

Order arithmetic: inequalities may be added and translated, and multiplying an inequality by a positive element preserves it; the order is total, so exactly one of a<1, a=1, a>1 holds (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)).

[L7]

Squeeze theorem; a constant sequence converges to its value; to establish convergence it suffices to produce a threshold for every real ε>0 (The squeeze theorem, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L8]

Algebra of limits, reciprocal rule: if zj→z with z≠0 and zj≠0 for every j, then 1/zj→1/z (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · cases
1.1

Let b be any real with b≥1 and let n≥1 be a natural. If b=1 then b1/n=1 and both inequalities hold. If b>1 then 1/n is a positive rational, so t:=b1/n−1>0; Bernoulli's inequality applied to t≥−1 gives b=(b1/n)n=(1+t)n≥1+ι(n)t, hence ι(n)t≤b−1 and t≤(b−1)/ι(n) since ι(n)>0. In both cases 1≤b1/n≤1+(b−1)/ι(n), which is claim 1.

givenL1L2L3L4L6
1.2

Case one: a=1.

givenassume-case one
1.3

Case big: a>1.

givenassume-case big
1.4

Case small: 0<a<1.

givenassume-case small
2.1

For every real b>1 the sequence b1/(k+1) converges to 1. Put dk:=1+(b−1)/ι(k+1). Given a real ε>0, the quotient ε/(b−1) is positive, so there is a natural p≥1 with 1/p<ε/(b−1); for k≥p we have k+1>p, hence ι(k+1)>ι(p)>0 and 0<(b−1)/ι(k+1)<(b−1)(1/p)<ε, so ∣dk−1∣<ε and dk→1. By step 1.1 applied at n=k+1 we have 1≤b1/(k+1)≤dk for every k, and the constant sequence 1 converges to 1, so the squeeze theorem gives b1/(k+1)→1.

step 1.1L4L5L6L7
2.2

In case one, ck=11/(k+1)=1 for every k, so (ck) is the constant sequence 1 and converges to 1.

step 1.2L1L7
3.1

In case big, a>1, so step 2.1 applied with b=a gives ck=a1/(k+1)→1.

step 2.1step 1.3
3.2

In case small, put a′:=1/a, which satisfies a′>1 because 0<a<1. For each natural n≥1 the product rule for roots gives a1/n(a′)1/n=(aa′)1/n=11/n=1, so a1/n=1/(a′)1/n, and (a′)1/n>0. By step 2.1 the sequence (a′)1/(k+1) converges to 1≠0 with all terms nonzero, so the reciprocal rule gives ck=1/(a′)1/(k+1)→1/1=1.

step 2.1step 1.4L1L2L5L8
4.1

The three cases are exhaustive by trichotomy applied to a and 1, the hypothesis a>0 excluding nothing else, and in each of them (ck) converges to 1; together with step 1.1 this proves both claims.

step 2.2step 3.1step 3.2step 1.1L6cases: trichotomy of the ordercases-exhaustive∎

Remarks

  • Bernoulli is doing the whole job in the case a>1. The inequality (1+t)n≥1+nt converts the exact identity (a1/n)n=a into the linear bound t≤(a−1)/n on the excess t=a1/n−1, and that bound is what tends to 0. No estimate on a1/n itself is needed beyond a1/n>1.

  • The case 0<a<1 is not symmetric to the case a>1 and is not proved again. It is transported by the reciprocal, using a1/n(1/a)1/n=1 (Laws of rational exponents) and the reciprocal rule of Algebra of limits: sums, scalar multiples, products and quotients. The hypothesis of that rule, that the limit be nonzero and every term nonzero, is met because roots of positive reals are positive.

  • The rate is different from the one in n1/n→1. Here the excess is O(1/n) with a constant depending on a; there the base itself grows with n and the excess is only O(1/n1/2). The two lemmas are therefore not instances of one another in either direction.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak

Statement

Let (ak)k∈N be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with ak>0 for every k. Put

qk:=ak+1ak,rk:=ak+11/(k+1)(k∈N),

with roots as in Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a and Rational powers ar of a positive base. Then, in R‾ (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined),

lim inf⁡kqk  ≤  lim inf⁡krk  ≤  lim sup⁡krk  ≤  lim sup⁡kqk.

The root sequence must start at index 1, and (rk) is the shift that makes it a sequence on N. The classical statement writes an1/n, which is meaningful only for n≥1, since 1/0 is not a rational number; sequences here are functions on N and N contains 0 (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so the root family is written rk=ak+11/(k+1), which is an1/n reindexed by n=k+1. The ratio family qk needs no shift, and the four quantities in the display are those of the two sequences (qk) and (rk) exactly as written here.

This is why the root test dominates the ratio test. If the ratios converge, the outer two quantities coincide and the chain forces the roots to converge to the same value; but the roots can converge when the ratios do not, and then the chain is strict at both ends. Both phenomena are exhibited by named examples on the companion page.

Facts & Assumptions

Given: A sequence (ak) of reals with ak>0 for every k; the ratio sequence qk=ak+1/ak; the root sequence rk=ak+11/(k+1); and ι(n)=n⋅1R for the canonical naturals.

[L2]

The order on R‾ is total and transitive, +∞ is greatest and −∞ least, it restricts on R to the order of R, and an element between two reals is real (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Epsilon characterisation, for a real L: L=lim sup⁡kzk gives zk<L+ε eventually for every real ε>0; L=lim inf⁡kzk gives zk>L−ε eventually for every real ε>0 (For finite L: L=lim sup⁡xk iff for every ε>0 one has xk<L+ε eventually and xk>L−ε frequently).

[L4]

lim inf⁡kzk≤lim sup⁡kzk (lim inf⁡xk≤lim sup⁡xk for every real sequence).

[L5]

Comparison: zk≤wk eventually implies lim sup⁡kzk≤lim sup⁡kwk and lim inf⁡kzk≤lim inf⁡kwk (If xk≤yk eventually then lim sup⁡xk≤lim sup⁡yk and lim inf⁡xk≤lim inf⁡yk).

[L6]

A sequence converging to a real c has lim sup⁡=lim inf⁡=c; and lim inf⁡kzk=+∞ implies zk→+∞, hence zk>M eventually for every real M (A real sequence converges to L∈R iff lim inf⁡xk=lim sup⁡xk=L, and diverges to ±∞ iff both equal ±∞, Divergence to +∞ and to −∞).

[L7]

For every real C>0 the sequence C1/(k+1) converges to 1 (For every a>0, a1/n→1).

[L8]

Algebra of limits: a scalar multiple of a convergent sequence converges to the scalar multiple of the limit (Algebra of limits: sums, scalar multiples, products and quotients).

[L9]

Roots and powers of positive reals: x1/n exists, is unique and is >0 for x>0 and n≥1; (xy)1/n=x1/ny1/n; the integer power xn is the rational power at exponent n, so (xn)1/n=xn⋅(1/n)=x; x−m=1/xm and xmxm′=xm+m′ for integer exponents and x≠0; xn>0 for x>0; and 0≤x≤y implies x1/n≤y1/n (Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Rational powers ar of a positive base, Laws of rational exponents, Monotonicity of r↦ar and of a↦ar, Integer powers am, Laws of integer exponents, Monotonicity of x↦xn and of n↦an).

[L10]
[L11]

Archimedean facts: for every real η>0 there is a natural m≥1 with 1/m<η; and 0<x<y gives 0<1/y<1/x (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L12]

Order arithmetic: Order is preserved by adding a constant and by adding inequalities and claim 4 of Sign rules for products and monotonicity of multiplication state the strict forms, that inequalities may be translated and added and that multiplication by a positive element preserves <; adjoining the case of equality, where both sides move or scale alike, gives the nonstrict forms used below. Products of nonnegative inequalities multiply in the nonstrict form stated by Multiplying inequalities of positives, and the order on R is total.

[L13]

Strictly between any two reals lies a rational (The rationals embed densely in the reals).

Proof

technique · direct
1.1

Every qk is positive, being a quotient of positive reals, and every rk is positive, being a root of the positive real ak+1. Hence 0 is a lower bound of every tail range of (qk) and of (rk), so every tail infimum is ≥0 and therefore lim inf⁡kqk≥0 and lim inf⁡krk≥0; with [L4] this also gives lim sup⁡kqk≥0.

givenL1L2L4L9L11
1.2

Let c>0 be real, let N∈N and put C:=aNc−N, a positive real. If ak+1≤c ak for every k≥N then an≤Ccn for every n≥N; if ak+1≥c ak for every k≥N then an≥Ccn for every n≥N. Both are inductions on j for n=N+j: at j=0 one has CcN=aNc−NcN=aNc0=aN, and the inductive step multiplies the bound at n by the positive c and uses the hypothesis at k=n.

givenL9L10L12
1.3

Let C>0 and c>0 be real and n≥1 a natural. Then (Ccn)1/n=C1/n(cn)1/n=C1/nc. Consequently 0<an≤Ccn gives an1/n≤C1/nc, and an≥Ccn>0 gives an1/n≥C1/nc, since x↦x1/n is nondecreasing on the nonnegative reals.

givenL9
1.4

For real C>0 and c>0 the sequence uk:=C1/(k+1)c converges to c, by [L7] and the scalar rule; hence lim sup⁡kuk=lim inf⁡kuk=c.

givenL6L7L8
1.5

If lim sup⁡kqk=+∞ then lim sup⁡krk≤lim sup⁡kqk, since +∞ is the greatest element of R‾.

givenL2
2.1

Suppose β:=lim sup⁡kqk is real, and let ε>0 be an arbitrary real. Put c:=β+ε, which is positive since β≥0. By [L3] there is N with qk<c for all k≥N, that is ak+1<c ak after multiplying by ak>0; so ak+1≤c ak for k≥N, and step 1.2 gives an≤Ccn for all n≥N with C:=aNc−N>0. For k≥N the index n:=k+1 satisfies n≥N and n≥1, so step 1.3 gives rk≤C1/(k+1)c=uk. By step 1.4 and [L5], lim sup⁡krk≤lim sup⁡kuk=c=β+ε.

step 1.1step 1.2step 1.3step 1.4L3L5L12L14
2.2

If α:=lim inf⁡kqk=0 then lim inf⁡krk≥0=α by step 1.1.

step 1.1
2.3

Suppose α:=lim inf⁡kqk>0 and let c be a real with 0<c<α. Then qk>c eventually: if α is real this is [L3] applied with ε:=α−c>0, and if α=+∞ then qk→+∞ by [L6], so qk>c eventually. Fix N with qk>c for all k≥N; then ak+1≥c ak for k≥N, so step 1.2 gives an≥Ccn for all n≥N with C:=aNc−N>0, and step 1.3 gives rk≥C1/(k+1)c=uk for every k≥N. By step 1.4 and [L5], lim inf⁡krk≥lim inf⁡kuk=c.

step 1.1step 1.2step 1.3step 1.4L3L5L6L12L14
3.1

Hence lim sup⁡krk≤lim sup⁡kqk. By step 1.1 the element β=lim sup⁡kqk is ≥0, so it is either +∞, which is step 1.5, or real. In the real case step 2.1 with ε=1 gives lim sup⁡krk≤β+1, a real, so lim sup⁡krk≠+∞; if lim sup⁡krk=−∞ it is ≤β; and otherwise it is a real S, and S>β would give, on choosing a natural m≥1 with 1/m<S−β and applying step 2.1 with ε=1/m, the impossibility S≤β+1/m<S. By totality lim sup⁡krk≤β.

step 2.1step 1.5step 1.1L2L11L12
3.2

Hence lim inf⁡kqk≤lim inf⁡krk. By step 1.1 the element α=lim inf⁡kqk is ≥0, so it is 0, or a positive real, or +∞. The first case is step 2.2. If α is a positive real and lim inf⁡krk<α, then lim inf⁡krk lies between the reals 0 and α by step 1.1 and is therefore real, so [L13] supplies a real c with lim inf⁡krk<c<α, necessarily c>0; step 2.3 then gives lim inf⁡krk≥c, contradicting c>lim inf⁡krk, so lim inf⁡krk≥α by totality. If α=+∞, step 2.3 gives lim inf⁡krk≥c for every real c with c>0, so lim inf⁡krk is not −∞, and it is not a real t either, since t≥0 by step 1.1 and then c:=t+1>0 would give t≥t+1; hence lim inf⁡krk=+∞=α.

step 2.2step 2.3step 1.1L2L12L13
4.1

Combining the three links, lim inf⁡kqk≤lim inf⁡krk by step 3.2, lim inf⁡krk≤lim sup⁡krk by [L4], and lim sup⁡krk≤lim sup⁡kqk by step 3.1.

step 3.1step 3.2L4∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

For every p>0 and every positive rational α, nα/(1+p)n→0

Statement

Let p∈R with p>0 and let α∈Q with α>0. Write ι(n):=n⋅1R for the canonical natural, with ι(0):=0, and let

wk  :=  ι(k)α(1+p)k(k∈N),

the numerator being a rational power (Rational powers ar of a positive base) and the denominator an integer power (Integer powers am). Then wk→0 (Limits and Cauchy sequences of reals).

Every term is defined, including the one at k=0. The supplementary clause of Rational powers ar of a positive base gives 0α=0 for rational α>0, and (1+p)0=1, so w0=0. No index shift is therefore needed here, in contrast with the two root lemmas earlier on this page, where the exponent is the index.

In words: a fixed power of n is beaten by any geometric sequence of ratio >1, however small the excess p and however large the exponent α.

Facts & Assumptions

Given: A real p>0 and a rational α>0; the base β:=1+p>1; the canonical naturals ι(n)=n⋅1R with ι(0)=0; and wk=ι(k)α/βk.

[L1]

Rational powers: xr is defined and positive for real x>0 and rational r, and 0r=0 for rational r>0; the integer power xm is the rational power at exponent m; (xy)r=xryr, which persists for x,y≥0 when r>0; x−r=1/xr; and (xr)s=xrs (Rational powers ar of a positive base, Laws of rational exponents, Integer powers am, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a).

[L2]

Monotonicity of rational powers: for rational t>0, x>1 implies xt>1; and for rational t>0, 0<x<y implies xt<yt (Monotonicity of r↦ar and of a↦ar).

[L3]

Integer powers: x>0 implies xm>0, and xmxm′=xm+m′, (xm)m′=xmm′ for integer exponents with x≠0 (Monotonicity of x↦xn and of n↦an, Laws of integer exponents).

[L4]

Bernoulli's inequality: (1+x)n≥1+ι(n)x for real x≥−1 and natural n (Bernoulli's inequality (1+x)n≥1+nx).

[L5]

Canonical naturals: ι(n)>0 and invertible for n≥1, and ι is strictly increasing (Canonical naturals are positive and strictly increasing, Order on the natural numbers, ≤ is a linear order on N).

[L6]

Reciprocal Archimedean property: for every real η>0 there is a natural m≥1 with 1/m<η; and 0<x<y gives 0<1/y<1/x (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L7]

Order arithmetic: Order is preserved by adding a constant and by adding inequalities and claim 4 of Sign rules for products and monotonicity of multiplication state the strict forms, that inequalities may be translated and added and that multiplication by a positive element preserves <; adjoining the case of equality gives the nonstrict forms used below. Products of nonnegative inequalities multiply in the nonstrict form stated by Multiplying inequalities of positives, and the order is total (Ordered field).

[L8]

Convergence to 0: it suffices to produce, for every real ε>0, a threshold beyond which ∣zk∣<ε; and ∣z∣=z for z≥0 (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Basic properties of the absolute value).

Proof

technique · direct
1.1

Since α>0 is rational, so is 1/α; put δ:=β1/α and θ:=δ1/2. From β>1 and 1/α>0 we get δ>1, and from δ>1 and 1/2>0 we get θ>1; hence θ−1>0 and δ>0, θ>0.

givenL1L2L7
1.2

For every natural n one has δn=θnθn, because θ2=(δ1/2)2=δ and therefore δn=(θ2)n=θ2n=θnθn.

givenL1L3
1.3

For every natural k one has wk=ukα, where uk:=ι(k)/δk. Indeed uk=ι(k)⋅(1/δk) with both factors ≥0, so ukα=ι(k)α(1/δk)α=ι(k)α/(δk)α, and (δk)α=δkα=(β1/α)kα=β(1/α)(kα)=βk.

givenL1L3
2.1

For every natural n≥1 one has 0≤un<1/(ι(n)(θ−1)(θ−1)). Bernoulli's inequality applied to θ−1>0 gives θn≥1+ι(n)(θ−1)>ι(n)(θ−1)>0, so multiplying this inequality by itself gives δn=θnθn>ι(n)(θ−1)ι(n)(θ−1)>0; dividing the positive ι(n) by the two positive quantities reverses the inequality and yields un=ι(n)/δn<ι(n)/(ι(n)ι(n)(θ−1)(θ−1))=1/(ι(n)(θ−1)(θ−1)), while un≥0 because ι(n)>0 and δn>0.

step 1.1step 1.2L3L4L5L6L7
3.1

The sequence (uk) converges to 0. Note first u0=ι(0)/δ0=0/1=0. Given a real ε>0, put η:=ε(θ−1)(θ−1)>0 and take a natural m≥1 with 1/m<η. For k≥m we have ι(k)≥ι(m)>0, hence 1/ι(k)≤1/ι(m)<η, and therefore 0≤uk<1/(ι(k)(θ−1)(θ−1))<η/((θ−1)(θ−1))=ε, so ∣uk∣<ε.

step 2.1L5L6L7L8
4.1

The sequence (wk) converges to 0. Given a real ε>0, the element ε1/α is a positive real, so by step 3.1 there is a threshold beyond which 0≤uk<ε1/α. For such k: if uk=0 then wk=0α=0<ε, and if uk>0 then monotonicity of the rational power α in the base gives wk=ukα<(ε1/α)α=ε(1/α)α=ε. In both cases ∣wk∣=wk<ε, so wk→0.

step 3.1step 1.3L1L2L8∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

For every real x, xk/k!→0

Statement

Write ι(n):=n⋅1R for the canonical natural (Canonical naturals are positive and strictly increasing) and define the factorial as the finite product (Finite sums and finite products, by recursion)

k!  :=  ∏j<kι(j+1)(k∈N),

so that 0!=1, the empty product, and (k+1)!=k!⋅ι(k+1). Every k! is a positive real. Then, for every x∈R,

xkk!⟶0,

the numerator being the integer power of Integer powers am and the convergence that of Limits and Cauchy sequences of reals.

The index range needs no adjustment: k! is defined at k=0 with value 1, and x0=1, so the sequence begins with x0/0!=1.

Facts & Assumptions

Given: A real x; the modulus M:=∣x∣≥0; the factorials k!=∏j<kι(j+1); and the canonical naturals ι(n)=n⋅1R.

[A1]

P(j) denotes the statement MN+j/(N+j)!≤Aλj, where N, λ and A are fixed in step 1.3.

[L1]

Finite products: the empty product is 1, ∏j<m+1aj=(∏j<maj)am, and a product of positive factors is positive (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

Integer powers: z0=1, zm+1=zmz, and z≥0 implies zm≥0 (Integer powers am, Monotonicity of x↦xn and of n↦an, Laws of integer exponents).

[L3]

Absolute value: ∣zw∣=∣z∣∣w∣, ∣z∣≥0, and ∣z∣=z for z≥0 (Basic properties of the absolute value).

[L4]
[L5]

Canonical naturals: ι(n)>0 and invertible for n≥1, ι is strictly increasing, and for every real y there is a natural N≥1 with y<ι(N) (Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean, Order on the natural numbers, ≤ is a linear order on N).

[L6]

Order arithmetic: Inverses of positives are positive, and reciprocation reverses order, claim 4 of Sign rules for products and monotonicity of multiplication and Order is preserved by adding a constant and by adding inequalities state the strict forms, that 0<u<v gives 0<1/v<1/u, that multiplication by a positive element preserves <, and that inequalities may be translated and added; adjoining the case of equality gives the nonstrict forms used below, and multiplication by 0 sends both sides to 0, so a nonnegative multiplier preserves ≤. Products of nonnegative inequalities multiply in the nonstrict form stated by Multiplying inequalities of positives.

[L7]

Geometric sequences: ∣r∣<1 implies rj→0 (For ∣r∣<1 the sequence rk is null, and for ∣r∣>1 the sequence ∣r∣k diverges to +∞); a scalar multiple of a convergent sequence converges to the scalar multiple of the limit (Algebra of limits: sums, scalar multiples, products and quotients).

[L8]

Squeeze theorem, and the fact that a constant sequence converges to its value (The squeeze theorem, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L9]

A sequence converges to z if and only if some tail of it does; the K-th tail of (zk) is j↦zj+K (Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · induction
1.1

Each k! is a product of the positive reals ι(j+1), j<k, hence positive, and (k+1)!=k!⋅ι(k+1); also M=∣x∣≥0.

givenL1L3L5
1.2

For every k∈N one has ∣xk∣=Mk: at k=0 both sides are ∣1∣=1=M0, and if ∣xk∣=Mk then ∣xk+1∣=∣xkx∣=∣xk∣∣x∣=MkM=Mk+1, so this follows by induction on k.

givenL2L3L4
1.3

Take a natural N≥1 with M<ι(N) and put λ:=M/ι(N) and A:=MN/N!. Then 0≤λ<1, since 0≤M<ι(N) and ι(N)>0, and A≥0.

givenL1L2L5L6choose
1.4

The statement P(0) holds, with equality: MN+0/(N+0)!=MN/N!=A=A⋅1=Aλ0.

givenA1L2base
1.5

Fix j∈N and assume P(j), that is MN+j/(N+j)!≤Aλj.

A1ih
2.1

Then P(j+1) holds. Indeed MN+j+1/(N+j+1)!=(MN+j/(N+j)!)(M/ι(N+j+1)), and N+j+1>N gives ι(N+j+1)>ι(N)>0, hence 0≤M/ι(N+j+1)≤M/ι(N)=λ; since also 0≤MN+j/(N+j)!≤Aλj by step 1.5 and Aλj≥0, multiplying the two nonnegative inequalities gives MN+j+1/(N+j+1)!≤Aλjλ=Aλj+1.

step 1.5A1L1L2L5L6
3.1

By the induction principle P(j) holds for every j∈N, and MN+j/(N+j)!≥0 always, so 0≤MN+j/(N+j)!≤Aλj for every j.

step 1.4step 2.1A1L1L2L4
4.1

Since ∣λ∣=λ<1, the sequence (λj)j converges to 0, hence so does (Aλj)j; the constant sequence 0 also converges to 0, so the squeeze theorem applied to step 3.1 shows that the N-th tail j↦MN+j/(N+j)! converges to 0, and therefore (Mk/k!)k converges to 0. Finally ∣xk/k!−0∣=∣xk∣/k!=Mk/k! by steps 1.1 and 1.2, so xk/k!→0.

step 3.1step 1.1step 1.2L3L7L8L9discharge-induction∎

Remarks

RemarkRemark: AI-generatedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Which extended-real operations this library leaves undefined, and where each lim sup⁡ statement needs the hypothesis

Conventions: sup⁡∅, unbounded sets, and the extended reals refused, inside R, the conventions sup⁡S=+∞ and inf⁡∅=+∞, and promised that a later page needing R‾ would introduce it explicitly as a new object with its own order and its own partial arithmetic. The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined is that introduction, and this page is the one that needed it. Nothing about the real supremum has changed: sup⁡S and inf⁡S for S⊆R are still real numbers, still defined only under the nonempty and bounded hypotheses, and the extended bounds of Every subset of R‾ has a least upper bound and a greatest lower bound in R‾, agreeing with the real supremum and infimum on nonempty sets bounded in R are a different operation in a different ordered set, agreeing with the real one exactly where the real one is defined.

What is defined. On R‾ this library defines exactly three things: the total order, the reflection a↦−a, and the two partial operations a+b and ab. The order and the reflection are total. The two operations are not, and the gaps are these.

expressionstatus
a+b with a,b∈Rthe field sum
(+∞)+b with b≠−∞+∞
(−∞)+b with b≠+∞−∞
(+∞)+(−∞) and (−∞)+(+∞)undefined
ab with a,b∈Rthe field product
(±∞)⋅b with b≠0±∞, by the sign rule
0⋅(±∞) and (±∞)⋅0undefined

What is not defined at all. There is no subtraction on R‾, no division, no absolute value and no exponentiation. Where a proof on this page wants a−b it writes a+(−b), which inherits the gap at {+∞,−∞}; where it wants a quotient it does not write one. In particular the expressions (+∞)−(+∞), (+∞)/(+∞) and 0/0 do not occur here, not because they are hard but because neither operation exists.

Why the two gaps are gaps. They are the two places where the value is not determined by the sequences involved, so no assignment could be compatible with limits. For the product this is proved: Null times divergent has no rule: xk=1/k with yk=ck gives product limit c, and with yk=k2 gives divergence ↗ exhibits a null sequence and sequences diverging to +∞ whose products behave differently, so 0⋅(+∞) has no value that would make a product rule true. For the sum the same is visible with ak=k and bk=−k, whose sum is constantly 0, against ak=k and bk=−2k, whose sum diverges to −∞; both pairs have lim sup⁡ak=+∞ and lim sup⁡bk=−∞.

Where each statement on this page carries the hypothesis. Reading the page in order, the pattern is that everything purely order-theoretic is unconditional and everything arithmetic is not.

What this costs a reader coming from a measure-theory text. Such a text typically declares 0⋅∞:=0, which is genuinely convenient there, because in an integral the factor 0 is a measure-zero set and the convention makes countable additivity work without cases. That convention is not in force here, and it is not compatible with limits: it is a decision about a particular formula, not a fact about R‾. A statement quoted from such a source therefore needs its degenerate cases restored before it can be used with the results on this page.

One thing that is not a convention. The equations lim sup⁡kxk=+∞ and lim inf⁡kxk=−∞ occurring throughout this page are ordinary equations between elements of R‾, not abbreviations. That is precisely the difference from Divergence to +∞ and to −∞, where "xk→+∞" is a single abbreviation for a condition and no object named +∞ is involved. Both readings coexist without conflict, and A real sequence converges to L∈R iff lim inf⁡xk=lim sup⁡xk=L, and diverges to ±∞ iff both equal ±∞ is the statement that relates them.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

FALSE: lim sup⁡(xk+yk)=lim sup⁡xk+lim sup⁡yk

Statement

False claim: for all sequences (xk), (yk) of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) whose limit superiors have a defined sum in R‾ (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined),

lim sup⁡k(xk+yk)  =  lim sup⁡kxk+lim sup⁡kyk.

The corresponding statement with = replaced by ≤ is true and is lim sup⁡(xk+yk)≤lim sup⁡xk+lim sup⁡yk whenever the right-hand side is defined in R‾, and dually for lim inf⁡. The claim above is what one gets by strengthening that inequality to an equality, and it fails: the two sides can differ by as much as the whole oscillation of the sequences, because the two limit superiors may be attained along different sets of indices while the sum of the sequences never sees either of them.

The witness is xk=(−1)k and yk=−(−1)k, refuted below; it is recorded separately as a named counterexample on the companion page.

Facts & Assumptions

[L2]

A strictly increasing index map satisfies nj≥j (A strictly increasing index map satisfies nk≥k).

[L4]

The order on R‾ is total and restricts on R to the order of R; ±∞ are not real (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

[L5]

Absolute value: ∣t∣=1 forces t=1 or t=−1 (Basic properties of the absolute value, Absolute value in an ordered field).

[L7]

Subadditivity: lim sup⁡k(zk+wk)≤lim sup⁡kzk+lim sup⁡kwk whenever the right-hand side is defined (lim sup⁡(xk+yk)≤lim sup⁡xk+lim sup⁡yk whenever the right-hand side is defined in R‾, and dually for lim inf⁡).

[L8]

The refuted claim: for all sequences whose limit superiors have a defined sum, lim sup⁡k(zk+wk)=lim sup⁡kzk+lim sup⁡kwk.

Refutation

technique · direct
1.1

The sequences xk=sk and yk=−sk are sequences of reals, and xk+yk=sk+(−sk)=0 for every k.

givenL1
1.2

Every value sk is 1 or −1, since ∣sk∣=1; and for every n both values occur at an index ≥n, since sen=1 with en≥n and son=−1 with on≥n.

givenL1L2L5
2.1

Hence Tn(x)={1,−1} for every n. Its least upper bound in R‾ is 1: the element 1 bounds both 1 and −1 from above because −1<1, and any upper bound u satisfies 1≤u because 1∈Tn(x). So sup⁡Tn(x)=1 for every n, and lim sup⁡kxk is the greatest lower bound of the one-element family {1}, namely 1.

step 1.2L3L4L6
2.2

The sequence yk=−sk takes the value 1 at every on and the value −1 at every en, and takes no other value, so Tn(y)={1,−1} for every n as well, and the same computation gives lim sup⁡kyk=1.

step 1.2L1L2L3L4L5L6
2.3

The sum sequence is constantly 0, so Tn(x+y)={0}, whose least upper bound is 0, and lim sup⁡k(xk+yk)=0.

step 1.1L3L4
3.1

Both limit superiors are the real number 1, so their sum is defined and equals 1+1=2, and the claim asserts 0=2 for this pair. But 0<2, so 0≠2 and the claim fails.

step 2.1step 2.2step 2.3L6L8
4.1

The claim is therefore false. What survives is the inequality of [L7], which for this pair reads 0≤2 and is strict.

step 3.1L7L8∎

Remarks

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak for every positive sequence

Statement

False claim: for every sequence (ak) of reals with ak>0 for all k,

lim sup⁡kak+11/(k+1)  =  lim sup⁡kak+1ak,

that is, the limit superior of the root sequence equals the limit superior of the ratio sequence. (The root family is written with the shift of For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak, since ak1/k is undefined at k=0; classically the claim reads lim sup⁡nan1/n=lim sup⁡nan+1/an.)

What is true is the chain lim inf⁡kak+1ak  ≤  lim inf⁡kak+11/(k+1)  ≤  lim sup⁡kak+11/(k+1)  ≤  lim sup⁡kak+1ak of For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak. The claim above collapses its right-hand inequality to an equality, and that fails: the roots can converge while the ratios oscillate. This is exactly why a root criterion decides cases that a ratio criterion cannot.

The witness is ak=2−k+(−1)k. The computation below establishes all four quantities for it, namely lim inf⁡kak+1ak=18,lim sup⁡kak+1ak=2,lim⁡kak+11/(k+1)=12, and it is recorded as a named example on the companion page.

Facts & Assumptions

Given: The alternating sequence (sk) and the index maps e,o of The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1; the sequence tk defined to be 2 when sk=1 and 1/2 when sk=−1; the sequence ak:=2−ktk; the ratios qk:=ak+1/ak and the roots rk:=ak+11/(k+1).

[L1]

The alternating sequence: s0=1, sk+1=−sk, ∣sk∣=1 for every k, sej=1 and soj=−1, and e, o are strictly increasing (The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1); a strictly increasing index map satisfies nj≥j (A strictly increasing index map satisfies nk≥k).

[L4]

Powers and roots of positive reals: integer powers with 2m2m′=2m+m′ and 2−m=1/2m; (xy)1/n=x1/ny1/n; the integer power is the rational power at an integer exponent, so (2−n)1/n=2−n/n=2−1; roots of positive reals are positive; and 0<x≤y implies x1/n≤y1/n (Integer powers am, Laws of integer exponents, Rational powers ar of a positive base, Laws of rational exponents, Monotonicity of r↦ar and of a↦ar, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a).

[L5]

For every real b>0 the sequence b1/(k+1) converges to 1 (For every a>0, a1/n→1).

[L9]

The refuted claim: for every sequence of positive reals, lim sup⁡kak+11/(k+1)=lim sup⁡kak+1/ak.

Refutation

technique · direct
1.1

Each sk is 1 or −1 because ∣sk∣=1, so tk is well defined, with tk∈{2,1/2} and tk>0; hence ak=2−ktk>0 for every k, and (ak) is a sequence of positive reals to which the claim applies. This is the sequence usually written ak=2−k+(−1)k.

givenL1L4L7L9
1.2

Since sk+1=−sk and 1≠−1, exactly one of the two situations "sk=1 and sk+1=−1" and "sk=−1 and sk+1=1" occurs at each index k. In the first, tk=2 and tk+1=1/2; in the second, tk=1/2 and tk+1=2.

givenL1L7
1.3

For every n both values of s occur at an index ≥n: sen=1 with en≥n and son=−1 with on≥n.

givenL1
2.1

The ratios are qk=ak+1/ak=(2−(k+1)tk+1)/(2−ktk)=2−1tk+1/tk, which by step 1.2 equals 2−1(1/2)/2=1/8 when sk=1 and 2−1⋅2/(1/2)=2 when sk=−1.

step 1.1step 1.2L4L7algebra
2.2

The roots are rk=(2−(k+1)tk+1)1/(k+1)=(2−(k+1))1/(k+1)tk+11/(k+1)=2−1tk+11/(k+1).

step 1.1L4
2.3

Since 1/2≤tk+1≤2 for every k and x↦x1/(k+1) is nondecreasing on the positive reals, (1/2)1/(k+1)≤tk+11/(k+1)≤21/(k+1); both bounding sequences converge to 1 by [L5], so the squeeze theorem gives tk+11/(k+1)→1.

step 1.1L4L5L6L7
3.1

By steps 1.2 and 1.3 the tail range of (qk) at every index n is exactly {1/8,2}: those are the only values, and each occurs at some index ≥n. Its least upper bound in R‾ is 2 and its greatest lower bound is 1/8, since 1/8<2 and both belong to the set; hence lim sup⁡kqk is the greatest lower bound of {2}, namely 2, and lim inf⁡kqk is the least upper bound of {1/8}, namely 1/8.

step 2.1step 1.3L2L7
3.2

By steps 2.2 and 2.3 and the scalar rule, rk=2−1tk+11/(k+1)→2−1⋅1=1/2, so lim sup⁡krk=lim inf⁡krk=1/2.

step 2.2step 2.3L3L6
4.1

For this sequence the claim asserts lim sup⁡krk=lim sup⁡kqk, that is 1/2=2; but 1/2<1<2, so the two are different and the claim fails.

step 3.1step 3.2L7L9
5.1

The claim is therefore false. The true chain [L8] reads here 1/8≤1/2≤1/2≤2, so both outer inequalities are strict for this witness while the middle one is an equality.

step 4.1step 3.1step 3.2L7L8L9∎

Remarks

Sources