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

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$a_{k} = 2^{- k + (- 1)^{k}}$ 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

1.1givenL1L3

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 $1 / 2$ .

1.2givenL4L5L6

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 $a_{k} \to 0$ .

2.1step 1.1L1L2L6

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.

3.1step 1.2step 2.1L1L6∎

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.

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.

$a_{k} = 2^{- k + (- 1)^{k}}$ 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

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Depends on

Dependency tree · 2 levels

Sources

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