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

1.1givenL1L5

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

1.2givenL1L2

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 $o_{n} \geq n$ .

2.1step 1.1step 1.2L3L4L6

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 $- 1$ .

3.1step 2.1L3L4

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 ${lim inf}_{k} s_{k} = - 1$ .

4.1step 3.1L6L7∎

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 $- \infty$ .

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 $lim sup (x_{k} y_{k}) = 0 < 4$ .

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

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

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