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✓ 9 results · all verified · 7 also independently AI-judged

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limsup, liminf, and Subsequential Limits: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

$(- 1)^{k}$ 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 $lim sup (x_{k} y_{k}) = 0 < 4$ .

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

The block sequence $1 / 1; 1 / 2, 2 / 2; 1 / 3, 2 / 3, 3 / 3; \dots$ has subsequential limit set exactly $[0, 1]$

Example

Write $ι (n) = n \cdot 1_{R}$ for the canonical natural. Let $c : N \to N \times N$ be the function supplied by the recursion theorem (The recursion theorem) from the starting element $(1, 1)$ and the function

$F (m, j) : = (m, j + 1) if j < m, F (m, j) : = (m + 1, 1) if j = m,$

write $c_{k} = (m_{k}, j_{k})$ , and define the sequence of reals

$x_{k} : = \frac{ι (j_{k})}{ι (m_{k})} (k \in N) .$

Its first terms are

$\frac{1}{1}; \frac{1}{2}, \frac{2}{2}; \frac{1}{3}, \frac{2}{3}, \frac{3}{3}; \frac{1}{4}, \dots$

the blocks being the successive values of $m$ . Then the subsequential limit set (Subsequential limit of a real sequence, and the subsequential limit set) is the whole unit interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length),

$SL (x) = [0, 1],$

and consequently ${lim sup}_{k} x_{k} = 1$ and ${lim inf}_{k} x_{k} = 0$ (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 recursion replaces the block bookkeeping. Presenting the sequence by the partial sums $1 + 2 + \dots + (m - 1)$ that mark where each block begins would require inverting that count at every index. Carrying the pair $(m, j)$ along instead makes each term's block and position immediately available, and the three facts the argument needs, that $1 \leq j_{k} \leq m_{k}$ , that $m_{k} \leq k + 1$ , and that every admissible pair occurs, are then three short inductions.

Facts & Assumptions

Given: The recursion $c_{k} = (m_{k}, j_{k})$ described above, the sequence $x_{k} = ι (j_{k}) / ι (m_{k})$ , and a real number $y$ with $0 \leq y \leq 1$ .

[L1]

Recursion theorem: for a set $A$ , an element $a \in A$ and $f : A \to A$ there is a unique $g : N \to A$ with $g_{0} = a$ and $g_{k + 1} = f (g_{k})$ (The recursion theorem, The natural numbers $N$ (von Neumann)).

[L2]

Induction principle (The principle of mathematical induction).

[L3]

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

[L4]

Index maps: if $n_{i} < n_{i + 1}$ for every $i$ then $n$ is strictly increasing, and then $n_{i} \geq i$ ; the composite $(x_{n_{i}})$ is a subsequence, and $L \in SL (x)$ means some subsequence converges to $L$ (A strictly increasing index map satisfies $n_{k} \geq k$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences, Subsequential limit of a real sequence, and the subsequential limit set, Limits and Cauchy sequences of reals).

[L5]

Canonical naturals: $ι (n) > 0$ and invertible for $n \geq 1$ , $ι$ is strictly increasing, $ι (1) = 1$ , and $ι (n + n^{'}) = ι (n) + ι (n^{'})$ (Canonical naturals are positive and strictly increasing).

[L6]

Order arithmetic: claim 4 of Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities and Inverses of positives are positive, and reciprocation reverses order state the strict forms, that multiplication by a positive element preserves $<$ , that inequalities may be translated and added, and that $0 < u < v$ gives $0 < 1 / v < 1 / u$ ; adjoining the case of equality, where the two sides coincide, 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, Complete ordered field (least-upper-bound property)).

[L7]

Reciprocal Archimedean property: for every real $η > 0$ there is a natural $p \geq 1$ with $1 / p < η$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L8]

Limits preserve non-strict inequalities (Limits preserve non-strict inequalities); a convergent sequence is bounded and a sequence diverging to $\pm \infty$ is unbounded (Every convergent sequence is bounded, Divergence to $+ \infty$ and to $- \infty$ ).

[L9]

The interval $[0, 1] = {t \in R : 0 \leq t \leq 1}$ , with least element $0$ and greatest element $1$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L10]

${lim sup}_{k} x_{k}$ is the greatest and ${lim inf}_{k} x_{k}$ the least element of $\overset{}{SL} (x)$ , whose real part is $SL (x)$ (The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one, The limit inferior is the least subsequential limit in $\overline{R}$ , Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, 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).

[L11]

Absolute value: $∣ t ∣ \leq η$ if and only if $- η \leq t \leq η$ ; and the order on $N$ is total with $k < k + 1$ (Basic properties of the absolute value, Absolute value in an ordered field, Order on the natural numbers, $\leq$ is a linear order on $N$ , Discreteness: $σ (n)$ is the immediate successor).

Verification

technique · constructive

1.1

The recursion theorem applied to $A = N \times N$ , the element $(1, 1)$ and the function $F$ gives a unique $c$ with $c_{0} = (1, 1)$ and $c_{k + 1} = F (c_{k})$ , so $(m_{k}, j_{k})$ and hence $x_{k}$ are well defined once $m_{k} \geq 1$ is known.

givenL1

1.2

For every $k$ one has $1 \leq j_{k} \leq m_{k}$ : this holds at $k = 0$ , where $(m_{0}, j_{0}) = (1, 1)$ ; and if $1 \leq j_{k} \leq m_{k}$ then either $j_{k} < m_{k}$ , so that $c_{k + 1} = (m_{k}, j_{k} + 1)$ with $1 \leq j_{k} + 1 \leq m_{k}$ , or $j_{k} = m_{k}$ , so that $c_{k + 1} = (m_{k} + 1, 1)$ with $1 \leq 1 \leq m_{k} + 1$ . The claim follows by induction on $k$ ; in particular $m_{k} \geq 1$ always, so $ι (m_{k}) > 0$ and $x_{k}$ is defined.

givenL2L5L11

1.3

For every $k$ one has $m_{k} \leq k + 1$ : at $k = 0$ , $m_{0} = 1$ ; and $m_{k + 1}$ is $m_{k}$ or $m_{k} + 1$ , so $m_{k + 1} \leq m_{k} + 1 \leq k + 2$ . This is again an induction on $k$ .

givenL2L11

1.4

For every natural $m \geq 1$ and every $j$ with $1 \leq j \leq m$ there is $k$ with $(m_{k}, j_{k}) = (m, j)$ . Induct on $m$ . For $m = 1$ the only admissible $j$ is $1$ , realised at $k = 0$ . Assume the claim for $m$ and apply it at $j = m$ to get $k$ with $(m_{k}, j_{k}) = (m, m)$ ; then $c_{k + 1} = (m + 1, 1)$ , and a second induction on $i$ shows $c_{k + 1 + i} = (m + 1, 1 + i)$ for every $i \leq m$ : it holds at $i = 0$ , and if it holds at $i < m$ then $1 + i < m + 1$ , so $c_{k + 2 + i} = (m + 1, 2 + i)$ . Hence every $j$ with $1 \leq j \leq m + 1$ is realised in block $m + 1$ .

givenL1L2L11

1.5

Let $y$ be a real with $0 \leq y \leq 1$ .

givenL9

2.1

For every $k$ one has $0 < x_{k} \leq 1$ : from $1 \leq j_{k} \leq m_{k}$ we get $0 < ι (j_{k}) \leq ι (m_{k})$ , and dividing by $ι (m_{k}) > 0$ gives $0 < x_{k} \leq 1$ .

step 1.2L5L6

2.2

For every natural $m \geq 1$ the set ${j \in N : 1 \leq j \leq m and y ι (m) \leq ι (j)}$ is nonempty, since $j = m$ belongs to it because $y \leq 1$ gives $y ι (m) \leq ι (m)$ ; let $J (m)$ be its least element. Then $y \leq ι (J (m)) / ι (m)$ , and moreover $ι (J (m)) / ι (m) \leq y + 1 / ι (m)$ : if $J (m) = 1$ this reads $1 / ι (m) \leq y + 1 / ι (m)$ , true because $y \geq 0$ ; and if $J (m) > 1$ then $j : = J (m) - 1$ satisfies $1 \leq j \leq m$ and $j < J (m)$ , so minimality gives $ι (j) < y ι (m)$ , that is $ι (J (m)) - 1 < y ι (m)$ , and dividing by $ι (m) > 0$ gives the bound. Hence $0 \leq ι (J (m)) / ι (m) - y \leq 1 / ι (m)$ .

step 1.5L3L5L6L11

3.1

For every $L \in SL (x)$ one has $0 \leq L \leq 1$ , since a subsequence of $(x_{k})$ converging to $L$ satisfies $0 \leq x_{n_{i}} \leq 1$ at every index by step 2.1, and limits preserve non-strict inequalities. So $SL (x) \subseteq [0, 1]$ .

step 2.1L4L8L9

3.2

For every natural $m \geq 1$ the set ${k \in N : (m_{k}, j_{k}) = (m, J (m))}$ is nonempty by step 1.4, since $1 \leq J (m) \leq m$ ; let $K (m)$ be its least element. Then $x_{K (m)} = ι (J (m)) / ι (m)$ , so $∣ x_{K (m)} - y ∣ \leq 1 / ι (m)$ by step 2.2, and $m = m_{K (m)} \leq K (m) + 1$ by step 1.3.

step 1.4step 1.3step 2.2L3L11

4.1

Define $f : N \to N$ by $f (n) : = K (n + 2)$ ; then $f (n) \geq (n + 2) - 1 = n + 1 > n$ and $∣ x_{f (n)} - y ∣ \leq 1 / ι (n + 2)$ . The recursion theorem applied to $N$ , the element $K (1)$ and the function $f$ gives $n : N \to N$ with $n_{0} = K (1)$ and $n_{i + 1} = f (n_{i})$ ; it is strictly increasing, so $n_{i} \geq i$ , and $∣ x_{n_{i + 1}} - y ∣ \leq 1 / ι (n_{i} + 2) \leq 1 / ι (i + 2)$ .

step 3.2L1L4L5L6construct

5.1

The subsequence $(x_{n_{i}})$ converges to $y$ : given a real $ε > 0$ , take a natural $p \geq 1$ with $1 / p < ε$ ; every $i \geq p$ satisfies $i \geq 1$ , so step 4.1 applied at $i - 1$ gives $∣ x_{n_{i}} - y ∣ \leq 1 / ι (i + 1)$ , and $i + 1 > p$ gives $1 / ι (i + 1) < 1 / p < ε$ . Hence $y \in SL (x)$ , and since $y$ was an arbitrary element of $[0, 1]$ , $[0, 1] \subseteq SL (x)$ .

step 4.1step 1.5L4L5L6L7L11

6.1

Therefore $SL (x) = [0, 1]$ . The sequence is bounded by step 2.1, so every subsequence of it is bounded and none diverges to $\pm \infty$ ; hence $\overset{}{SL} (x) = SL (x) = [0, 1]$ , whose greatest element is $1$ and least element $0$ , and [L10] gives ${lim sup}_{k} x_{k} = 1$ and ${lim inf}_{k} x_{k} = 0$ .

step 5.1step 3.1step 2.1L8L9L10discharge-construct∎

Remarks

Every point of $[0, 1]$ is approached, and the rate is the block width. In block $m$ the terms are $1 / m, 2 / m, \dots, m / m$ , spaced $1 / m$ apart and covering $(0, 1]$ , so any target in $[0, 1]$ has a term of block $m$ within $1 / m$ of it. Since blocks of every width occur, and occur arbitrarily late, this produces a subsequence converging to the target. That is the whole idea; steps 2.2 and 3.2 only make the choice of term canonical, by taking a least element rather than an arbitrary one, so that no choice principle is used.
The set is closed, as it must be. $[0, 1]$ contains the limit of every convergent sequence of its own points, which is what If each $y_{j}$ is a subsequential limit of $(x_{k})$ and $y_{j} \to y \in R$ , then $y$ is a subsequential limit of $(x_{k})$ predicts for any subsequential limit set. This example shows the prediction is not vacuous: the set here is an entire interval, in contrast with the two-point set of $(- 1)^{k}$ has $lim inf = - 1$ and $lim sup = 1$ , so it does not converge.
Neither endpoint is a value of the sequence in the case of $0$ . Every term is $> 0$ by step 2.1, so $0 \in SL (x)$ is a genuine limit and not an attained value, while $1$ is attained, once in every block. Subsequential limits need not be values, and values need not be subsequential limits.
Why the sequence is not written by a closed formula. The classical presentation defines $x_{k}$ by first solving $m (m - 1) / 2 \leq k < m (m + 1) / 2$ for $m$ , which needs a least-element argument at every index and a quadratic estimate to get $m \leq k + 1$ . The recursion carries the block and position forward instead, and the estimate $m_{k} \leq k + 1$ becomes the one-line induction of step 1.3.

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

$a_{k} = 2^{- k + (- 1)^{k}}$ has $lim inf a_{k + 1} / a_{k} = 1 / 8$ , $lim sup a_{k + 1} / a_{k} = 2$ and $\lim a_{k}^{1 / k} = 1 / 2$

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 $s \circ o \equiv - 1$ , let $t_{k} : = 2$ when $s_{k} = 1$ and $t_{k} : = 1 / 2$ when $s_{k} = - 1$ , and put

$a_{k} : = 2^{- k} t_{k} (k \in N),$

the sequence usually written $a_{k} = 2^{- k + (- 1)^{k}}$ . Writing $q_{k} : = a_{k + 1} / a_{k}$ for the ratios and $r_{k} : = a_{k + 1}^{1 / (k + 1)}$ for the roots, which is $a_{n}^{1 / n}$ reindexed by $n = k + 1$ as For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ requires,

${lim inf}_{k} q_{k} = \frac{1}{8}, {lim sup}_{k} q_{k} = 2, \lim_{k} r_{k} = \frac{1}{2},$

so also ${lim inf}_{k} r_{k} = {lim sup}_{k} r_{k} = 1 / 2$ . In addition $a_{k} \to 0$ .

The point. The ratios oscillate across $1$ , taking the values $1 / 8$ and $2$ alternately, so no statement of the form "the ratios are eventually below some $λ < 1$ " is available; the roots, by contrast, converge to $1 / 2 < 1$ . Any criterion reading the ratios alone is silent here, and one reading the roots is not. That is the concrete form of the dominance recorded in For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ .

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ , the auxiliary $t_{k} \in {2, 1 / 2}$ , the sequence $a_{k} = 2^{- k} t_{k}$ , the ratios $q_{k} = a_{k + 1} / a_{k}$ and the roots $r_{k} = a_{k + 1}^{1 / (k + 1)}$ , all as in FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence.

[L1]

For this sequence: every $a_{k}$ is positive, $q_{k} \in {1 / 8, 2}$ with both values occurring at arbitrarily large indices, ${lim inf}_{k} q_{k} = 1 / 8$ , ${lim sup}_{k} q_{k} = 2$ , and $r_{k} \to 1 / 2$ , so ${lim inf}_{k} r_{k} = {lim sup}_{k} r_{k} = 1 / 2$ (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence).

[L2]

The chain ${lim inf}_{k} q_{k} \leq {lim inf}_{k} r_{k} \leq {lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k}$ (For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ ).

[L3]

Limit superior and limit inferior in $\overline{R}$ (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 extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L4]

Powers: $2^{- k} = (1 / 2)^{k}$ and $t_{k} \leq 2$ , so $a_{k} \leq 2 \cdot (1 / 2)^{k}$ ; and $a_{k} > 0$ (Integer powers $a^{m}$ , Laws of integer exponents, Rational powers $a^{r}$ of a positive base, Sign rules for products and monotonicity of multiplication).

[L5]

Geometric sequences: $∣ ρ ∣ < 1$ implies $ρ^{k} \to 0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ); the squeeze theorem and the scalar rule (The squeeze theorem, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

Order arithmetic: $0 < 1$ , so $1 / 8 < 1 / 2 < 1 < 2$ and $1 / 2 \neq 2$ ; $∣ t ∣ = 1$ forces $t = \pm 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Basic properties of the absolute value, Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)).

Verification

technique · direct

1.1

The three values in the display are exactly what [L1] records for this sequence, together with ${lim inf}_{k} r_{k} = {lim sup}_{k} r_{k} = 1 / 2$ , which follows from the convergence of $(r_{k})$ to $1 / 2$ .

givenL1L3

1.2

The sequence is null: $0 < a_{k} = 2^{- k} t_{k} \leq 2 \cdot (1 / 2)^{k}$ for every $k$ , and $(1 / 2)^{k} \to 0$ because $∣ 1 / 2 ∣ < 1$ , so $2 \cdot (1 / 2)^{k} \to 0$ and the squeeze gives $a_{k} \to 0$ .

givenL4L5L6

2.1

The ratio quantities differ from one another and from the root quantities: $1 / 8 < 1 / 2 < 2$ , so ${lim inf}_{k} q_{k} < {lim inf}_{k} r_{k} = {lim sup}_{k} r_{k} < {lim sup}_{k} q_{k}$ . In particular the chain [L2] holds here with both outer inequalities strict and the middle one an equality, and the ratios do not determine the roots.

step 1.1L1L2L6

3.1

So $(a_{k})$ is a positive null sequence whose root sequence converges to $1 / 2 < 1$ while its ratio sequence has ${lim sup}_{k} q_{k} = 2 > 1$ and ${lim inf}_{k} q_{k} = 1 / 8 < 1$ , that is, the ratios oscillate across $1$ while the roots settle strictly below it.

step 1.2step 2.1L1L6∎

Remarks

Where the numbers come from. The exponent $- k + (- 1)^{k}$ changes by $- 1 + (- 1)^{k + 1} - (- 1)^{k} = - 1 \mp 2$ from one index to the next, giving ratios $2^{- 3} = 1 / 8$ and $2^{1} = 2$ ; the root divides the exponent by the index, so the bounded oscillation contributes $2^{\pm 1 / (k + 1)} \to 1$ and only the linear part $- k$ survives, giving $2^{- 1} = 1 / 2$ . The full computation is in FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence.
The same sequence reappears for series. With these $a_{k}$ the series $\sum_{k} a_{k}$ converges, and the root criterion sees it while the ratio criterion does not. That use belongs to the series page and is not made here.
Strictness of the middle inequality needs a different witness. Here ${lim inf}_{k} r_{k} = {lim sup}_{k} r_{k}$ , since the roots converge. A sequence making all three inequalities of the chain strict is A positive sequence making all three inequalities of the ratio-to-root chain strict.

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

A positive sequence making all three inequalities of the ratio-to-root chain strict

Example

$a_{k} : = 2^{- k} when s_{k} = 1, a_{k} : = 3^{- k} when s_{k} = - 1.$

This interleaves the two geometric sequences $2^{- k}$ and $3^{- k}$ , taking the first at even indices and the second at odd ones. With $q_{k} : = a_{k + 1} / a_{k}$ and $r_{k} : = a_{k + 1}^{1 / (k + 1)}$ as in For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ ,

${lim inf}_{k} q_{k} = 0, {lim inf}_{k} r_{k} = \frac{1}{3}, {lim sup}_{k} r_{k} = \frac{1}{2}, {lim sup}_{k} q_{k} = + \infty,$

so the chain of that theorem reads

$0 < \frac{1}{3} < \frac{1}{2} < + \infty$

with all three inequalities strict.

Where each comparison lives. The first two, $0 < 1 / 3$ and $1 / 3 < 1 / 2$ , are comparisons of real numbers and hold in $R$ ; they hold in $\overline{R}$ as well only because the extended order 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). The third, $1 / 2 < + \infty$ , is not a comparison in $R$ at all: $+ \infty$ is not a real number, and the inequality is the instance of "every real is below the greatest element" in $\overline{R}$ . So the outer two values of the chain are of different kinds here, and only the extended line can hold all four at once.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ with index maps $e, o$ (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 $a_{k}$ defined above; the ratios $q_{k} = a_{k + 1} / a_{k}$ ; and the roots $r_{k} = a_{k + 1}^{1 / (k + 1)}$ .

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ , $s_{k + 1} = - s_{k}$ , $s_{e_{j}} = 1$ , $s_{o_{j}} = - 1$ , with $e$ , $o$ strictly increasing, so $e_{j} \geq j$ and $o_{j} \geq j$ ; also $o_{0} = σ (0) \geq 1$ (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 $s \circ o \equiv - 1$ , A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Limit superior and limit inferior in $\overline{R}$ : existence for every sequence, the tail supremum being the least upper bound of the tail range and the tail infimum its 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).

[L3]

The order on $\overline{R}$ is total, $+ \infty$ is greatest, every real is $< + \infty$ , and the order 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).

[L4]

Powers: $2^{- k} = (1 / 2)^{k}$ and $3^{- k} = (1 / 3)^{k}$ ; $x^{m} x^{m^{'}} = x^{m + m^{'}}$ and $(x y)^{m} = x^{m} y^{m}$ for integer exponents and nonzero bases; $x^{m} > 0$ for $x > 0$ ; $(x^{- n})^{1 / n} = x^{- 1}$ for $x > 0$ and $n \geq 1$ (Integer powers $a^{m}$ , Laws of integer exponents, Rational powers $a^{r}$ of a positive base, Laws of rational exponents, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[L5]

Geometric sequences: $∣ ρ ∣ < 1$ implies $ρ^{k} \to 0$ , and $∣ ρ ∣ > 1$ implies $∣ ρ ∣^{k} \to + \infty$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ , Limits and Cauchy sequences of reals, Divergence to $+ \infty$ and to $- \infty$ ).

[L6]

Order arithmetic: $0 < 1$ , so $0 < 1 / 3 < 1 / 2 < 1 < 2 < 3$ ; multiplying an inequality by a positive element preserves it; reciprocals reverse the order; the order is total; $∣ t ∣ = 1$ forces $t = 1$ or $t = - 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value, Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)).

[L7]

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

[L8]

Verification

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ , so $a_{k}$ is well defined, and $a_{k} > 0$ for every $k$ since positive powers of positive bases are positive.

givenL1L4L6

1.2

For every $n \in N$ there are indices $k, k^{'} \geq n$ with $s_{k} = 1$ and $s_{k^{'}} = - 1$ , namely $k = e_{n}$ and $k^{'} = o_{n}$ ; and there are indices $l, l^{'} \geq n$ with $s_{l + 1} = 1$ and $s_{l^{'} + 1} = - 1$ , namely $l = e_{j} - 1$ and $l^{'} = o_{j} - 1$ for any $j \geq n + 1$ , these being natural numbers because $e_{j} \geq j \geq 1$ and $o_{j} \geq j \geq 1$ , and satisfying $l \geq j - 1 \geq n$ and $l^{'} \geq j - 1 \geq n$ .

givenL1L7

1.3

Since $s_{k + 1} = - s_{k}$ , the ratios are $q_{k} = 3^{- (k + 1)} / 2^{- k} = 3^{- 1} (2 / 3)^{k}$ when $s_{k} = 1$ , and $q_{k} = 2^{- (k + 1)} / 3^{- k} = 2^{- 1} (3 / 2)^{k}$ when $s_{k} = - 1$ ; in both cases $q_{k} > 0$ .

givenL1L4L6

1.4

Likewise the roots are $r_{k} = (2^{- (k + 1)})^{1 / (k + 1)} = 2^{- 1}$ when $s_{k + 1} = 1$ , and $r_{k} = (3^{- (k + 1)})^{1 / (k + 1)} = 3^{- 1}$ when $s_{k + 1} = - 1$ .

givenL1L4

2.1

By steps 1.2 and 1.4 the tail range of $(r_{k})$ at every index $n$ is exactly ${1 / 2, 1 / 3}$ , whose least upper bound is $1 / 2$ and greatest lower bound $1 / 3$ , since $1 / 3 < 1 / 2$ and both belong to the set. Hence ${lim sup}_{k} r_{k} = 1 / 2$ and ${lim inf}_{k} r_{k} = 1 / 3$ .

step 1.2step 1.4L2L3L6

2.2

${lim sup}_{k} q_{k} = + \infty$ . Fix $n$ and a real $M$ . Since $∣ 3 / 2 ∣ > 1$ , the sequence $(3 / 2)^{k}$ diverges to $+ \infty$ , so there is $K$ with $2^{- 1} (3 / 2)^{k} > M$ for all $k \geq K$ ; taking $j$ at least as large as both $n$ and $K$ and putting $k : = o_{j}$ , we get $k \geq j \geq n$ and $s_{k} = - 1$ , hence $q_{k} = 2^{- 1} (3 / 2)^{k} > M$ . So no real bounds the tail range of $(q_{k})$ above, its least upper bound in $\overline{R}$ is $+ \infty$ for every $n$ , and ${lim sup}_{k} q_{k}$ is the greatest lower bound of ${+ \infty}$ , namely $+ \infty$ .

step 1.2step 1.3L2L3L5L6L7

2.3

${lim inf}_{k} q_{k} = 0$ . Fix $n$ . All $q_{k}$ are positive, so $0$ is a lower bound of the tail range. If $ℓ > 0$ were a lower bound, then, since $∣ 2 / 3 ∣ < 1$ gives $(2 / 3)^{k} \to 0$ and hence $3^{- 1} (2 / 3)^{k} < ℓ$ for all $k \geq K$ for some $K$ , taking $j$ at least as large as both $n$ and $K$ and putting $k : = e_{j}$ would give $k \geq j \geq n$ , $s_{k} = 1$ and $q_{k} = 3^{- 1} (2 / 3)^{k} < ℓ$ , contradicting that $ℓ$ is a lower bound. So every lower bound is $\leq 0$ and the greatest lower bound of each tail range is $0$ ; hence ${lim inf}_{k} q_{k}$ is the least upper bound of ${0}$ , namely $0$ .

step 1.2step 1.3L2L3L5L6L7

3.1

Collecting the four values, the chain [L8] reads $0 \leq 1 / 3 \leq 1 / 2 \leq + \infty$ , and each inequality is strict: $0 < 1 / 3$ and $1 / 3 < 1 / 2$ hold in $R$ and therefore in $\overline{R}$ , while $1 / 2 < + \infty$ holds because $+ \infty$ is the greatest element of $\overline{R}$ and is distinct from every real. So no two of the four quantities coincide.

step 2.1step 2.2step 2.3L3L6L8∎

Remarks

Why interleaving two different geometric sequences does it. Each root is determined by the base used at a single index, so the root sequence takes only the two values $1 / 2$ and $1 / 3$ , and its limit superior and limit inferior are those two numbers. Each ratio, by contrast, compares two different bases at consecutive indices, so it contains a factor $(2 / 3)^{k}$ or $(3 / 2)^{k}$ and runs to $0$ on one subsequence and to $+ \infty$ on the other. Widening the gap between the two bases widens the outer two values without moving the inner two.
Compare $a_{k} = 2^{- k + (- 1)^{k}}$ has $lim inf a_{k + 1} / a_{k} = 1 / 8$ , $lim sup a_{k + 1} / a_{k} = 2$ and $\lim a_{k}^{1 / k} = 1 / 2$ . There the roots converge, so the middle inequality is an equality and only the outer two are strict. Here all three are strict, which is the most that For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ permits.
Both outer values are attained by the ratios in the extreme sense. The ratio sequence has $lim inf = 0$ and $lim sup = + \infty$ , so the ratio data place no restriction whatever on the roots beyond the chain, and the chain is therefore the sharpest general statement relating the two.

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

The four standard limits $n^{1 / n} \to 1$ , $a^{1 / n} \to 1$ , $n^{α} / (1 + p)^{n} \to 0$ and $x^{k} / k! \to 0$ , computed

Example

The four standard limits of this page, written as sequences on $N$ and instantiated. Throughout $ι (n) = n \cdot 1_{R}$ is the canonical natural, with $ι (0) = 0$ .

classical form	as a sequence on $N$	value	source
$n^{1 / n} \to 1$	$(k + 1)^{1 / (k + 1)}$	$1$	$n^{1 / n} \to 1$
$a^{1 / n} \to 1$ , $a > 0$	$a^{1 / (k + 1)}$	$1$	For every $a > 0$ , $a^{1 / n} \to 1$
$n^{α} / (1 + p)^{n} \to 0$	$ι (k)^{α} / (1 + p)^{k}$	$0$	For every $p > 0$ and every positive rational $α$ , $n^{α} / (1 + p)^{n} \to 0$
$x^{k} / k! \to 0$	$x^{k} / k!$	$0$	For every real $x$ , $x^{k} / k! \to 0$

Two of the four need an index shift and two do not, and the reason is visible in the classical forms: in the first two the index sits in the exponent as $1 / n$ , which is not a rational number at $n = 0$ , so those families begin at $n = 1$ and are written here with $n = k + 1$ . In the last two the index sits in the base or in a factorial, both of which are defined at $0$ , so no shift is needed and the sequences begin at $k = 0$ with the values $0$ and $1$ respectively.

The instances computed below are:

$2^{1 / (k + 1)} \to 1, \frac{ι (k)^{2}}{2^{k}} \to 0, \frac{2^{k}}{k!} \to 0, \frac{(- 3)^{k}}{k!} \to 0, \frac{ι (k)^{2}}{k!} \to 0.$

The last of these is not one of the four; it is the composite that orders the three scales, and it is obtained from two of them by the product rule.

Facts & Assumptions

Given: The canonical naturals $ι (n) = n \cdot 1_{R}$ with $ι (0) = 0$ ; the factorial $k! = \prod_{j < k} ι (j + 1)$ of For every real $x$ , $x^{k} / k! \to 0$ ; rational powers (Rational powers $a^{r}$ of a positive base) and integer powers (Integer powers $a^{m}$ ).

[L1]

$(k + 1)^{1 / (k + 1)} \to 1$ , with $1 \leq n^{1 / n} \leq 1 + 2 / n^{1 / 2}$ for every natural $n \geq 1$ ( $n^{1 / n} \to 1$ ).

[L2]

For every real $a > 0$ , $a^{1 / (k + 1)} \to 1$ ; and for real $b \geq 1$ and natural $n \geq 1$ , $1 \leq b^{1 / n} \leq 1 + (b - 1) / ι (n)$ (For every $a > 0$ , $a^{1 / n} \to 1$ ).

[L3]

For every real $p > 0$ and rational $α > 0$ , $ι (k)^{α} / (1 + p)^{k} \to 0$ (For every $p > 0$ and every positive rational $α$ , $n^{α} / (1 + p)^{n} \to 0$ ).

[L4]

For every real $x$ , $x^{k} / k! \to 0$ (For every real $x$ , $x^{k} / k! \to 0$ ).

[L5]

Algebra of limits: products of convergent sequences converge to the product of the limits (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

Arithmetic: $0 < 1 < 2 < 3$ , so $1$ , $2$ , $3 / 2$ and $1 / 2$ are positive and $2 = 1 + 1$ , $3 / 2 = 1 + 1 / 2$ ; a positive integer power of a positive real is positive and nonzero; $2$ is a positive rational and $1 / 2$ is a positive rational (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Canonical naturals are positive and strictly increasing, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Finite sums and finite products, by recursion, Ordered field, Complete ordered field (least-upper-bound property)).

Verification

technique · direct

1.1

The first standard limit gives $(k + 1)^{1 / (k + 1)} \to 1$ , together with the explicit two-sided bound $1 \leq (k + 1)^{1 / (k + 1)} \leq 1 + 2 / (k + 1)^{1 / 2}$ valid at every $k \in N$ , since $k + 1 \geq 1$ .

givenL1

1.2

The second gives $2^{1 / (k + 1)} \to 1$ and $(1 / 2)^{1 / (k + 1)} \to 1$ , both bases being positive; for the first of these the explicit bound of [L2] with $b = 2$ reads $1 \leq 2^{1 / n} \leq 1 + 1 / ι (n)$ for every natural $n \geq 1$ .

givenL2L6

1.3

The third, with $p = 1$ and $α = 2$ , gives $ι (k)^{2} / 2^{k} \to 0$ ; with $p = 1 / 2$ and $α = 1 / 2$ it gives $ι (k)^{1 / 2} / (3 / 2)^{k} \to 0$ . Both $p$ are positive reals and both $α$ are positive rationals, so [L3] applies in each case.

givenL3L6

1.4

The fourth, with $x = 2$ and with $x = - 3$ , gives $2^{k} / k! \to 0$ and $(- 3)^{k} / k! \to 0$ ; no hypothesis on $x$ is needed, in particular no positivity.

givenL4

2.1

Multiplying the first limit of step 1.3 by the first of step 1.4 and using $2^{k} \neq 0$ gives $ι (k)^{2} / k! = (ι (k)^{2} / 2^{k}) (2^{k} / k!) \to 0 \cdot 0 = 0$ .

step 1.3step 1.4L5L6

2.2

Multiplying the limit of step 1.1 by the first of step 1.2 gives $(k + 1)^{1 / (k + 1)} 2^{1 / (k + 1)} \to 1 \cdot 1 = 1$ .

step 1.1step 1.2L5

3.1

The four limits and the two composites are therefore established as displayed, and together they order the three growth scales: a fixed power of $n$ is beaten by every geometric sequence of ratio $> 1$ by step 1.3, every geometric sequence is beaten by the factorial by step 1.4, and consequently a fixed power of $n$ is beaten by the factorial by step 2.1.

step 1.1step 1.2step 1.3step 1.4step 2.1step 2.2∎

Remarks

The two bounds quoted in steps 1.1 and 1.2 are the useful part in practice. They convert the qualitative statement into a rate: $n^{1 / n}$ is within $2 / n^{1 / 2}$ of $1$ , and $a^{1 / n}$ within $(a - 1) / n$ of $1$ when $a \geq 1$ . The second rate is faster, and the difference is real: in $n^{1 / n}$ the base itself grows with the index.
Why $α$ is rational and $p$ is real. The exponent $α$ must be rational because rational powers are all this library has; the base $1 + p$ may be any real $> 1$ because it is raised only to integer powers. The asymmetry is a fact about what has been constructed, not about the mathematics, and it disappears once real exponents are available.
The composite in step 2.1 is the one usually quoted as "factorials beat polynomials". It is not proved directly anywhere on this page: it is the product of two of the four standard limits, and the product rule (Algebra of limits: sums, scalar multiples, products and quotients) is what assembles it.
Nothing here uses $lim sup$ . All four are ordinary limits, and the page's machinery is needed only to prove them, not to state them; the connection to the rest of the page is that For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ is the tool that makes several of them routine once one of them is known.

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

$x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$

Statement refuted

That the subadditivity of $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ can be improved to an equality: that for all sequences $(x_{k})$ , $(y_{k})$ of reals whose limit superiors have a defined sum, ${lim sup}_{k} (x_{k} + y_{k}) = {lim sup}_{k} x_{k} + {lim sup}_{k} y_{k}$ . The claim is recorded and refuted as FALSE: $lim sup (x_{k} + y_{k}) = lim sup x_{k} + lim sup y_{k}$ ; this item is the named witness.

Facts & Assumptions

[L1]

For this pair of sequences: $x_{k} + y_{k} = 0$ for every $k$ , ${lim sup}_{k} x_{k} = 1$ , ${lim sup}_{k} y_{k} = 1$ and ${lim sup}_{k} (x_{k} + y_{k}) = 0$ (FALSE: $lim sup (x_{k} + y_{k}) = lim sup x_{k} + lim sup y_{k}$ ).

[L2]

Subadditivity: ${lim sup}_{k} (x_{k} + y_{k}) \leq {lim sup}_{k} x_{k} + {lim sup}_{k} y_{k}$ whenever the right-hand side is defined ( $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ ).

[L3]

Limit superior in $\overline{R}$ , and the fact that a sum of two real numbers is the field sum (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 extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Order arithmetic: $0 < 1$ , hence $0 < 1 + 1 = 2$ and $0 \neq 2$ (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)).

Counterexample

technique · direct

1.1

The two sequences $x_{k} = s_{k}$ and $y_{k} = - s_{k}$ are sequences of reals whose termwise sum is constantly $0$ , and their limit superiors are both the real number $1$ .

givenL1L3

1.2

Both limit superiors being real, their sum is defined in $\overline{R}$ and equals the field sum $1 + 1 = 2$ .

givenL1L3L4

2.1

The limit superior of the sum sequence is $0$ , while the sum of the limit superiors is $2$ , and $0 < 2$ ; so the inequality of [L2] holds here and is strict, and the equality asserted by the refuted claim fails.

step 1.1step 1.2L1L2L4∎

Remarks

The gap is as large as the oscillation. Each sequence oscillates between $- 1$ and $1$ , and the two oscillations are exactly out of phase, so the sum never sees either peak. Replacing $y$ by a sequence in phase with $x$ , for instance $y = x$ , restores equality; the failure is a statement about the relative phase, not about the sizes involved.
Both hypotheses of $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ hold here. The two limit superiors are real, so their sum is defined, and no boundedness is needed for that theorem at all. The counterexample therefore refutes the equality on the theorem's own terms, not by exploiting a degenerate case.
The dual statement fails in the dual way. For the same pair, ${lim inf}_{k} x_{k} = {lim inf}_{k} y_{k} = - 1$ while ${lim inf}_{k} (x_{k} + y_{k}) = 0$ , so the $lim inf$ inequality ${lim inf}_{k} (x_{k} + y_{k}) \geq {lim inf}_{k} x_{k} + {lim inf}_{k} y_{k}$ of $lim sup (x_{k} + y_{k}) \leq lim sup x_{k} + lim sup y_{k}$ whenever the right-hand side is defined in $\overline{R}$ , and dually for $lim inf$ is also strict here, reading $0 \geq - 2$ .

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

$x_{k} = 1 + (- 1)^{k}$ , $y_{k} = 1 + (- 1)^{k + 1}$ give $lim sup (x_{k} y_{k}) = 0 < 4$

Statement refuted

That the submultiplicativity of For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$ can be improved to an equality: that for all bounded nonnegative sequences $(x_{k})$ , $(y_{k})$ of reals, ${lim sup}_{k} (x_{k} y_{k}) = ({lim sup}_{k} x_{k}) ({lim sup}_{k} y_{k}) .$

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ 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 index maps $e, o$ ; the sequences $x_{k} : = 1 + s_{k}$ and $y_{k} : = 1 - s_{k}$ , which are the families usually written $1 + (- 1)^{k}$ and $1 + (- 1)^{k + 1}$ ; and the tail ranges $T_{n}$ of 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 $s_{o_{j}} = - 1$ , and $e$ , $o$ are strictly increasing with $n_{j} \geq j$ (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 $s \circ o \equiv - 1$ , A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Limit superior in $\overline{R}$ : existence for every sequence, and the least-upper-bound and greatest-lower-bound descriptions of the tail bounds (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, The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L3]

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

[L4]

Order and field arithmetic: $0 < 1$ , so $0 < 2 = 1 + 1$ and $0 < 4 = 2 \cdot 2$ ; $1 + 1 = 2$ , $1 - 1 = 0$ , $1 + (- 1) = 0$ and $1 - (- 1) = 2$ ; a product with a zero factor is $0$ (The multiplicative identity is positive, 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)).

[L5]

Submultiplicativity: for bounded nonnegative sequences, ${lim sup}_{k} (x_{k} y_{k}) \leq ({lim sup}_{k} x_{k}) ({lim sup}_{k} y_{k})$ , all three quantities being real (For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$ ).

Counterexample

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ . When $s_{k} = 1$ the pair $(x_{k}, y_{k})$ is $(2, 0)$ , and when $s_{k} = - 1$ it is $(0, 2)$ . In either case $0 \leq x_{k} \leq 2$ and $0 \leq y_{k} \leq 2$ , so both sequences are bounded and nonnegative, and $x_{k} y_{k} = 0$ because one of the two factors is $0$ .

givenL1L3L4

1.2

For every $n$ both cases occur at an index $\geq n$ : $s_{e_{n}} = 1$ with $e_{n} \geq n$ and $s_{o_{n}} = - 1$ with $o_{n} \geq n$ .

givenL1

2.1

Hence $T_{n} (x) = {0, 2}$ and $T_{n} (y) = {0, 2}$ for every $n$ , each with least upper bound $2$ in $\overline{R}$ , since $2$ bounds both elements and belongs to the set; so ${lim sup}_{k} x_{k} = {lim sup}_{k} y_{k} = 2$ . The product sequence is constantly $0$ , so $T_{n} (x y) = {0}$ and ${lim sup}_{k} (x_{k} y_{k}) = 0$ .

step 1.1step 1.2L2L4

3.1

The hypotheses of [L5] are met by step 1.1, and the inequality it gives reads $0 \leq 2 \cdot 2 = 4$ . Since $0 < 4$ , it is strict, so the equality asserted above fails for this pair and the refuted claim is false.

step 2.1step 1.1L4L5∎

Remarks

The two sequences vanish at complementary indices. Wherever $x_{k}$ attains its maximum $2$ , its partner $y_{k}$ is $0$ , so the product is $0$ everywhere and the two limit superiors are attained along disjoint sets of indices. This is the multiplicative form of the phase mismatch behind $x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$ .
Nonnegativity and boundedness are satisfied, so the failure is not degenerate. Both hypotheses of For bounded nonnegative sequences, $lim sup (x_{k} y_{k}) \leq (lim sup x_{k}) (lim sup y_{k})$ hold, and the right-hand side is an honest product of real numbers; the inequality simply cannot be reversed.
The gap can be made total. Here the product sequence is identically zero while the bound is $4$ , so no fraction of the bound is achieved. Equality does hold when one of the two sequences converges, for the same reason as in the additive case.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-08-08 (gpt-5.6-terra-codex-subscription)Open item page →

A sequence with $lim sup = + \infty$ : the greatest subsequential limit exists only in $\overline{R}$

Statement refuted

That The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one can be stated inside $R$ : that for every sequence $(x_{k})$ of reals the set $SL (x)$ of real subsequential limits (Subsequential limit of a real sequence, and the subsequential limit set) has a greatest element and that element is ${lim sup}_{k} x_{k}$ .

The witness below has a nonempty $SL (x)$ with a greatest element, so the failure is not that the real set is empty: it is that the greatest element of $SL (x)$ is $0$ while ${lim sup}_{k} x_{k} = + \infty$ . The dominant behaviour of the sequence is invisible to $SL (x)$ and is recorded only by $\overset{}{SL} (x)$ (Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ ).

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 $s \circ o \equiv - 1$ ; the canonical naturals $ι (k) = k \cdot 1_{R}$ with $ι (0) = 0$ ; and the sequence $x_{k} : = ι (k)$ when $s_{k} = 1$ and $x_{k} : = 0$ when $s_{k} = - 1$ .

[L1]

The alternating sequence: $∣ s_{k} ∣ = 1$ , $s_{e_{j}} = 1$ , $s_{o_{j}} = - 1$ , and $e$ , $o$ are strictly increasing, so $e_{j} \geq j$ and $o_{j} \geq j$ (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 $s \circ o \equiv - 1$ , A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Limit superior in $\overline{R}$ : existence for every sequence, the tail supremum being the least upper bound of the tail range and $lim sup$ the greatest lower bound of the family of tail suprema (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).

[L3]

The order on $\overline{R}$ is total, $+ \infty$ is greatest, every real is $< + \infty$ and $> - \infty$ , and the order 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).

[L4]

Extended subsequential limits and convergence in $\overline{R}$ ; divergence to $+ \infty$ means that for every real $M$ one has $x_{k} > M$ eventually (Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ , Divergence to $+ \infty$ and to $- \infty$ , Subsequential limit of a real sequence, and the subsequential limit set, Limits and Cauchy sequences of reals).

[L5]

Canonical naturals: $ι$ is strictly increasing with $ι (k) \geq 0$ , and for every real $M$ there is a natural $p \geq 1$ with $M < ι (p)$ (Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean).

[L6]

A convergent sequence of reals is bounded, a limit is unique, and a sequence agreeing with a constant from some index on converges to that constant (Every convergent sequence is bounded, A sequence has at most one limit, Convergence depends only on the tail, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Absolute value and order: $∣ t ∣ = 1$ forces $t = 1$ or $t = - 1$ ; $0 < 1$ ; the order on $N$ is total (Basic properties of the absolute value, Absolute value in an ordered field, The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Order on the natural numbers, $\leq$ is a linear order on $N$ , Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ , so $(x_{k})$ is a well-defined sequence of reals with $x_{k} \geq 0$ for every $k$ ; moreover $x_{e_{j}} = ι (e_{j})$ and $x_{o_{j}} = 0$ for every $j$ .

givenL1L5L7

1.2

The subsequence along $o$ is constantly $0$ , and $o$ is strictly increasing, so $0 \in SL (x)$ .

givenL1L4L6

2.1

For every $n \in N$ the tail supremum $\sup T_{n} (x)$ is $+ \infty$ . Given a real $M$ , take a natural $p \geq 1$ with $M < ι (p)$ and an index $j$ at least as large as both $n$ and $p$ ; then $e_{j} \geq j \geq n$ , so $x_{e_{j}} = ι (e_{j}) \in T_{n} (x)$ , and $e_{j} \geq j \geq p$ gives $ι (e_{j}) \geq ι (p) > M$ . So no real number bounds $T_{n} (x)$ above, and the least upper bound in $\overline{R}$ must be $+ \infty$ .

step 1.1L1L2L3L5L7

3.1

Every real subsequential limit of $(x_{k})$ equals $0$ . Let $n$ be strictly increasing with $x_{n_{i}} \to L \in R$ ; the subsequence is then bounded, say $∣ x_{n_{i}} ∣ \leq B$ for every $i$ . Suppose $s_{n_{i}} = 1$ for arbitrarily large $i$ : taking a natural $p \geq 1$ with $B < ι (p)$ and such an index $i \geq p$ , we get $x_{n_{i}} = ι (n_{i}) \geq ι (i) \geq ι (p) > B$ , contradicting the bound. So there is $I$ with $s_{n_{i}} = - 1$ , hence $x_{n_{i}} = 0$ , for every $i \geq I$ ; a sequence equal to $0$ from an index on converges to $0$ , so $L = 0$ by uniqueness of limits.

step 1.1step 2.1L1L4L5L6L7

4.1

Consequently ${lim sup}_{k} x_{k}$ is the greatest lower bound of the family ${+ \infty}$ , namely $+ \infty$ , while $SL (x) = {0}$ by steps 1.2 and 3.1, whose greatest element is the real number $0$ . Since $0 \neq + \infty$ , the refuted claim fails for this sequence.

step 2.1step 1.2step 3.1L2L3∎

Remarks

What the extended set records. By The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one the element ${lim sup}_{k} x_{k} = + \infty$ lies in $\overset{}{SL} (x)$ and is its greatest element, so $\overset{}{SL} (x) = {0, + \infty}$ : the value $- \infty$ is excluded because $x_{k} \geq 0$ for every $k$ , so no subsequence can be eventually below a negative real. The real set $SL (x) = {0}$ is exactly the finite part of it, as Convergence in $\overline{R}$ and the extended subsequential limit set: $L \in \overline{R}$ is an extended subsequential limit when some subsequence converges to $L$ , or diverges to $L = \pm \infty$ says it must be.
Why the theorem cannot simply be restricted to bounded sequences. For a bounded sequence $lim sup$ is real and the two statements agree; the point of stating The limit superior is itself a subsequential limit in $\overline{R}$ and is the greatest one in $\overline{R}$ is that it then holds for every sequence, with no hypothesis to check, and this witness shows the hypothesis-free version is strictly stronger.
A simpler witness would prove less. The sequence $x_{k} = ι (k)$ has $SL (x) = \emptyset$ , so the refuted claim fails there only because an empty set has no greatest element. Interleaving with $0$ makes $SL (x)$ nonempty with a greatest element, so the claim fails for the substantive reason: the greatest real subsequential limit is not the limit superior.

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-08-08 (gpt-5.6-terra-codex-subscription)Open item page →

Null times divergent has no rule: $x_{k} = 1 / k$ with $y_{k} = c k$ gives product limit $c$ , and with $y_{k} = k^{2}$ gives divergence

Statement refuted

That the product $0 \cdot (+ \infty)$ , left undefined by The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, could be given a value compatible with limits: that there is $v \in \overline{R}$ such that for all sequences of reals with $x_{k} \to 0$ (Limits and Cauchy sequences of reals) and $y_{k} \to + \infty$ (Divergence to $+ \infty$ and to $- \infty$ ) the products $x_{k} y_{k}$ have the single limiting behaviour named by $v$ .

Equivalently: that knowing a factor is null and the other diverges to $+ \infty$ determines anything at all about the product. It does not, and the two undefined entries in the arithmetic of $\overline{R}$ are undefined for exactly this reason.

Facts & Assumptions

Given: The canonical naturals $ι (n) = n \cdot 1_{R}$ ; the sequence $x_{k} : = 1 / ι (k + 1)$ ; for a real $c > 0$ the sequence $y_{k}^{(c)} : = c ι (k + 1)$ ; and the sequence $z_{k} : = ι (k + 1) ι (k + 1)$ .

[L1]

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

[L2]

Archimedean facts: for every real $η > 0$ there is a natural $p \geq 1$ with $1 / p < η$ , and for every real $M$ there is a natural $p \geq 1$ with $M < ι (p)$ ; and $0 < u < v$ gives $0 < 1 / v < 1 / u$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L3]

Convergence to a real and divergence to $+ \infty$ ; to establish convergence it suffices to produce a threshold for every real $ε > 0$ ; a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Divergence to $+ \infty$ and to $- \infty$ ).

[L4]

A sequence diverging to $+ \infty$ is unbounded and therefore does not converge to any real (Divergence to $+ \infty$ and to $- \infty$ , Every convergent sequence is bounded); a limit, when it exists, is unique (A sequence has at most one limit).

[L5]

Order and field arithmetic: multiplying an inequality by a positive element preserves it; $0 < 1 < 2$ and $1 \neq 2$ ; $u \cdot (1 / u) = 1$ for $u \neq 0$ ; and the algebra of limits (Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities, The multiplicative identity is positive, Algebra of limits: sums, scalar multiples, products and quotients, Integer powers $a^{m}$ , Ordered field, Complete ordered field (least-upper-bound property)).

[L6]

The product $0 \cdot (\pm \infty)$ is left undefined in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

Counterexample

technique · direct

1.1

The sequence $x_{k} = 1 / ι (k + 1)$ is well defined, positive, and converges to $0$ : given a real $ε > 0$ , take a natural $p \geq 1$ with $1 / p < ε$ ; for $k \geq p$ we have $ι (k + 1) > ι (p) > 0$ , hence $0 < x_{k} < 1 / p < ε$ .

givenL1L2L3

1.2

For every real $c > 0$ the sequence $y_{k}^{(c)} = c ι (k + 1)$ diverges to $+ \infty$ : given a real $M$ , the quotient $M / c$ is real, so there is a natural $p \geq 1$ with $M / c < ι (p)$ , and for $k \geq p$ we get $ι (k + 1) > ι (p) > M / c$ , hence $y_{k}^{(c)} = c ι (k + 1) > M$ after multiplying by $c > 0$ .

givenL1L2L3L5

1.3

The sequence $z_{k} = ι (k + 1) ι (k + 1)$ diverges to $+ \infty$ : given a real $M$ , take a natural $p \geq 1$ with $M < ι (p)$ ; for $k \geq p$ we have $ι (k + 1) \geq 1$ and $ι (k + 1) > ι (p) > M$ , so $z_{k} \geq ι (k + 1) > M$ .

givenL1L2L3L5

2.1

For every real $c > 0$ the product sequence is constant: $x_{k} y_{k}^{(c)} = (1 / ι (k + 1)) c ι (k + 1) = c$ for every $k$ , so it converges to $c$ .

step 1.1step 1.2L3L5

2.2

The product with $(z_{k})$ is $x_{k} z_{k} = (1 / ι (k + 1)) ι (k + 1) ι (k + 1) = ι (k + 1)$ , which diverges to $+ \infty$ by the argument of step 1.3 with the single factor, and therefore converges to no real number.

step 1.1step 1.3L1L2L3L4L5

3.1

Now take the three pairs $(x, y^{(1)})$ , $(x, y^{(2)})$ and $(x, z)$ . In each, the first sequence is null and the second diverges to $+ \infty$ , so each pair satisfies the hypotheses of the refuted claim; but the three products converge to $1$ , converge to $2$ , and converge to $+ \infty$ in the extended sense. Since $1 \neq 2$ and limits are unique, no single $v \in \overline{R}$ describes all three, and the claim is false.

step 2.1step 2.2L4L5L6∎

Remarks

This is why the entry is blank in the table. The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined leaves $0 \cdot (\pm \infty)$ undefined not out of caution but because any value assigned to it would make some instance of a product rule false, and the three pairs above already realise three different behaviours.
The same phenomenon rules out $(+ \infty) + (- \infty)$ . Taking $a_{k} = ι (k)$ and $b_{k} = - ι (k)$ gives a sum that is constantly $0$ , while $a_{k} = ι (k)$ and $b_{k} = - ι (k) - ι (k)$ gives a sum diverging to $- \infty$ ; both pairs have $a_{k} \to + \infty$ and $b_{k} \to - \infty$ .
Measure theory's convention is not a counterexample to this. Texts that set $0 \cdot \infty : = 0$ are fixing the value of a formula in a context where the factor $0$ is the measure of a null set, not asserting a limit rule; the distinction is spelled out in Which extended-real operations this library leaves undefined, and where each $lim sup$ statement needs the hypothesis.
Index range. The classical statement writes $x_{k} = 1 / k$ and $y_{k} = c k$ , which requires $k \geq 1$ . Written on $N$ , which contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the same sequences are $1 / (k + 1)$ and $c (k + 1)$ , as above.

limsup, liminf, and Subsequential Limits: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

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

Example

Facts & Assumptions

Verification

Remarks

The block sequence 1/1; 1/2,2/2; 1/3,2/3,3/3; …1/1; 1/2,2/2; 1/3,2/3,3/3; … has subsequential limit set exactly [0,1][0,1]

Example

Facts & Assumptions

Verification

Remarks

ak=2−k+(−1)kak​=2−k+(−1)k has lim inf⁡ak+1/ak=1/8liminfak+1​/ak​=1/8, lim sup⁡ak+1/ak=2limsupak+1​/ak​=2 and lim⁡ak1/k=1/2limak1/k​=1/2

Example

Facts & Assumptions

Verification

Remarks

A positive sequence making all three inequalities of the ratio-to-root chain strict

Example

Facts & Assumptions

Verification

Remarks

The four standard limits n1/n→1n1/n→1, a1/n→1a1/n→1, nα/(1+p)n→0nα/(1+p)n→0 and xk/k!→0xk/k!→0, computed

Example

Facts & Assumptions

Verification

Remarks

xk=(−1)kxk​=(−1)k, yk=(−1)k+1yk​=(−1)k+1 give lim sup⁡(xk+yk)=0<2=lim sup⁡xk+lim sup⁡yklimsup(xk​+yk​)=0<2=limsupxk​+limsupyk​

Statement refuted

Facts & Assumptions

Counterexample

Remarks

xk=1+(−1)kxk​=1+(−1)k, yk=1+(−1)k+1yk​=1+(−1)k+1 give lim sup⁡(xkyk)=0<4limsup(xk​yk​)=0<4

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A sequence with lim sup⁡=+∞limsup=+∞: the greatest subsequential limit exists only in R‾R

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Null times divergent has no rule: xk=1/kxk​=1/k with yk=ckyk​=ck gives product limit cc, and with yk=k2yk​=k2 gives divergence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

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

The block sequence $1 / 1; 1 / 2, 2 / 2; 1 / 3, 2 / 3, 3 / 3; \dots$ has subsequential limit set exactly $[0, 1]$

$a_{k} = 2^{- k + (- 1)^{k}}$ has $lim inf a_{k + 1} / a_{k} = 1 / 8$ , $lim sup a_{k + 1} / a_{k} = 2$ and $\lim a_{k}^{1 / k} = 1 / 2$

The four standard limits $n^{1 / n} \to 1$ , $a^{1 / n} \to 1$ , $n^{α} / (1 + p)^{n} \to 0$ and $x^{k} / k! \to 0$ , computed

$x_{k} = (- 1)^{k}$ , $y_{k} = (- 1)^{k + 1}$ give $lim sup (x_{k} + y_{k}) = 0 < 2 = lim sup x_{k} + lim sup y_{k}$

$x_{k} = 1 + (- 1)^{k}$ , $y_{k} = 1 + (- 1)^{k + 1}$ give $lim sup (x_{k} y_{k}) = 0 < 4$

A sequence with $lim sup = + \infty$ : the greatest subsequential limit exists only in $\overline{R}$

Null times divergent has no rule: $x_{k} = 1 / k$ with $y_{k} = c k$ gives product limit $c$ , and with $y_{k} = k^{2}$ gives divergence