ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$a_k = 2^{-k + (-1)^k}$ has $\liminf a_{k+1}/a_k = 1/8$ , $\limsup a_{k+1}/a_k = 2$ 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 $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , 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

$a_k \;:=\; 2^{-k} t_k \qquad (k \in \mathbb{N}),$

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 $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ requires,

$\liminf_{k} q_k = \frac{1}{8}, \qquad \limsup_{k} q_k = 2, \qquad \lim_{k} r_k = \frac{1}{2},$

so also $\liminf_k r_k = \limsup_k r_k = 1/2$ . 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 $\lambda < 1$ " 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 $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ .

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: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ 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, $\liminf_k q_k = 1/8$ , $\limsup_k q_k = 2$ , and $r_k \to 1/2$ , so $\liminf_k r_k = \limsup_k r_k = 1/2$ (FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ for every positive sequence).

[L2]

The chain $\liminf_k q_k \le \liminf_k r_k \le \limsup_k r_k \le \limsup_k q_k$ (For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ ).

[L3]

Limit superior and limit inferior in $\overline{\mathbb{R}}$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined).

[L4]

Powers: $2^{-k} = (1/2)^{k}$ and $t_k \le 2$ , so $a_k \le 2 \cdot (1/2)^{k}$ ; 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: $|\rho| < 1$ implies $\rho^{k} \to 0$ (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 $1/2 \ne 2$ ; $|t| = 1$ 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 $\liminf_k r_k = \limsup_k r_k = 1/2$ , which follows from the convergence of $(r_k)$ to $1/2$ .

givenL1L3

1.2

The sequence is null: $0 < a_k = 2^{-k} t_k \le 2 \cdot (1/2)^{k}$ 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 $\liminf_k q_k < \liminf_k r_k = \limsup_k r_k < \limsup_k q_k$ . 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 $\limsup_k q_k = 2 > 1$ and $\liminf_k q_k = 1/8 < 1$ , 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: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ 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 $\liminf_k r_k = \limsup_k r_k$ , 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources

Ratio test (Wikipedia) (standard reference, not scraped)
Root test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.35) (standard reference, not scraped)

ak=2−k+(−1)ka_k = 2^{-k + (-1)^k}ak​=2−k+(−1)k has lim inf⁡ak+1/ak=1/8\liminf a_{k+1}/a_k = 1/8liminfak+1​/ak​=1/8, lim sup⁡ak+1/ak=2\limsup a_{k+1}/a_k = 2limsupak+1​/ak​=2 and lim⁡ak1/k=1/2\lim a_k^{1/k} = 1/2limak1/k​=1/2