ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$(- 1)^{k}$ has $lim inf = - 1$ and $lim sup = 1$ , so it does not converge

Example

Let $(s_{k})$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , the sequence usually written $s_{k} = (- 1)^{k}$ , characterised by $s_{0} = 1$ and $s_{k + 1} = - s_{k}$ . Then

${lim inf}_{k} s_{k} = - 1, {lim sup}_{k} s_{k} = 1,$

so the two differ and $(s_{k})$ neither converges nor diverges to $\pm \infty$ (A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , and diverges to $\pm \infty$ iff both equal $\pm \infty$ ).

This is the smallest example in which the inequality $lim inf \leq lim sup$ is strict, and it shows exactly what the gap measures: the sequence keeps returning to two different values, and neither of them can be the limit because the other keeps interrupting.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ 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 $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , and the tail ranges $T_{n} = {s_{k} : k \geq n}$ with extended tail bounds $s ⁣ \sup_{n} : = \sup T_{n}$ and $s ⁣ \inf_{n} : = \inf T_{n}$ (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ ).

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ for every $j$ , 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 $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

A strictly increasing index map satisfies $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L3]

Limit superior and limit inferior: ${lim sup}_{k} z_{k} = \inf {\sup T_{n} (z)}$ and ${lim inf}_{k} z_{k} = \sup {\inf T_{n} (z)}$ , all four kinds of bound existing in $\overline{R}$ for every sequence, the supremum being the least upper bound and the infimum the greatest lower bound (Limit superior and limit inferior of a real sequence as $\inf_{n} \sup_{k \geq n} x_{k}$ and $\sup_{n} \inf_{k \geq n} x_{k}$ in $\overline{R}$ , The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , 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, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

The order on $\overline{R}$ is total and restricts on $R$ to the order of $R$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , 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).

[L6]

Order arithmetic: $0 < 1$ , hence $- 1 < 1$ and $- 1 \neq 1$ (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)).

[L7]

A real sequence converges to $L \in R$ exactly when $lim inf = lim sup = L$ , and diverges to $\pm \infty$ exactly when both equal $\pm \infty$ (A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , and diverges to $\pm \infty$ iff both equal $\pm \infty$ ).

Verification

technique · direct

1.1

Every value of the sequence is $1$ or $- 1$ , since $∣ s_{k} ∣ = 1$ .

givenL1L5

1.2

For every $n \in N$ both values occur at some index $\geq n$ : $s_{e_{n}} = 1$ with $e_{n} \geq n$ , and $s_{o_{n}} = - 1$ with $o_{n} \geq n$ .

givenL1L2

2.1

Hence $T_{n} = {1, - 1}$ for every $n$ . Its least upper bound in $\overline{R}$ is $1$ , since $1$ bounds both elements from above, using $- 1 < 1$ , and any upper bound is $\geq 1$ because $1 \in T_{n}$ ; dually its greatest lower bound is $- 1$ .

step 1.1step 1.2L3L4L6

3.1

Therefore the family of tail suprema is the one-element family ${1}$ , whose greatest lower bound is $1$ , so ${lim sup}_{k} s_{k} = 1$ ; and the family of tail infima is ${- 1}$ , whose least upper bound is $- 1$ , so ${lim inf}_{k} s_{k} = - 1$ .

step 2.1L3L4

4.1

Since $- 1 \neq 1$ , there is no $L$ with ${lim inf}_{k} s_{k} = {lim sup}_{k} s_{k} = L$ , so by [L7] the sequence converges to no real number and diverges to neither $+ \infty$ nor $- \infty$ .

step 3.1L6L7∎

Remarks

The two values are exactly the subsequential limits. By The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one and The limit inferior is the least subsequential limit in $\overline{R}$ the numbers $1$ and $- 1$ are the greatest and least elements of $\overset{}{SL} (s)$ , and since every term is $1$ or $- 1$ no other value can be a subsequential limit, so $SL (s) = {- 1, 1}$ exactly.
Contrast with The sequence $(- 1)^{k} (1 + 1 / k)$ is bounded with subsequential limit set exactly ${- 1, 1}$ . There the same two subsequential limits arise for a sequence none of whose terms equals either of them. The limit superior and limit inferior do not care: they are determined by the tails, not by whether the values are attained.
This sequence is the standard witness for strictness throughout the page. It drives FALSE: $lim sup (x_{k} + y_{k}) = lim sup x_{k} + lim sup y_{k}$ and $x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$ , and, after an affine change, $x_{k} = 1 + (- 1)^{k}$ , $y_{k} = 1 + (- 1)^{k + 1}$ give $lim sup (x_{k} y_{k}) = 0 < 4$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Limit superior and limit inferior (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

(−1)k(−1)k has lim inf⁡=−1liminf=−1 and lim sup⁡=1limsup=1, so it does not converge