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

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FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ for every positive sequence

Statement

False claim: for every sequence $(a_k)$ of reals with $a_k > 0$ for all $k$ ,

$\limsup_{k} a_{k+1}^{1/(k+1)} \;=\; \limsup_{k} \frac{a_{k+1}}{a_k},$

that is, the limit superior of the root sequence equals the limit superior of the ratio sequence. (The root family is written with the shift of For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ , since $a_k^{1/k}$ is undefined at $k = 0$ ; classically the claim reads $\limsup_n a_n^{1/n} = \limsup_n a_{n+1}/a_n$ .)

What is true is the chain $\liminf_{k} \frac{a_{k+1}}{a_k} \;\le\; \liminf_{k} a_{k+1}^{1/(k+1)} \;\le\; \limsup_{k} a_{k+1}^{1/(k+1)} \;\le\; \limsup_{k} \frac{a_{k+1}}{a_k}$ of For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ . The claim above collapses its right-hand inequality to an equality, and that fails: the roots can converge while the ratios oscillate. This is exactly why a root criterion decides cases that a ratio criterion cannot.

The witness is $a_k = 2^{-k + (-1)^k}$ . The computation below establishes all four quantities for it, namely $\liminf_{k} \frac{a_{k+1}}{a_k} = \frac{1}{8}, \qquad \limsup_{k} \frac{a_{k+1}}{a_k} = 2, \qquad \lim_{k} a_{k+1}^{1/(k+1)} = \frac{1}{2},$ and it is recorded as a named example on the companion page.

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 $\mathbb{N}$ 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 $s \circ o \equiv -1$ ; the sequence $t_k$ defined to be $2$ when $s_k = 1$ and $1/2$ when $s_k = -1$ ; the sequence $a_k := 2^{-k} t_k$ ; the ratios $q_k := a_{k+1}/a_k$ and the roots $r_k := a_{k+1}^{1/(k+1)}$ .

[L1]

The alternating sequence: $s_0 = 1$ , $s_{k+1} = -s_k$ , $|s_k| = 1$ for every $k$ , $s_{e_j} = 1$ and $s_{o_j} = -1$ , and $e$ , $o$ are strictly increasing (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 $s \circ o \equiv -1$ ); a strictly increasing index map satisfies $n_j \ge j$ (A strictly increasing index map satisfies $n_k \ge k$ ).

[L2]

Limit superior and limit inferior in $\overline{\mathbb{R}}$ , their 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 $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set, The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined).

[L3]

A sequence converging to a real $c$ has $\limsup = \liminf = c$ (A real sequence converges to $L \in \mathbb{R}$ iff $\liminf x_k = \limsup x_k = L$ , and diverges to $\pm\infty$ iff both equal $\pm\infty$ ).

[L4]

Powers and roots of positive reals: integer powers with $2^{m} 2^{m'} = 2^{m+m'}$ and $2^{-m} = 1/2^{m}$ ; $(xy)^{1/n} = x^{1/n} y^{1/n}$ ; the integer power is the rational power at an integer exponent, so $\big(2^{-n}\big)^{1/n} = 2^{-n/n} = 2^{-1}$ ; roots of positive reals are positive; and $0 < x \le y$ implies $x^{1/n} \le y^{1/n}$ (Integer powers $a^m$ , Laws of integer exponents, Rational powers $a^r$ of a positive base, Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ ).

[L5]

For every real $b > 0$ the sequence $b^{1/(k+1)}$ converges to $1$ (For every $a > 0$ , $a^{1/n} \to 1$ ).

[L6]

Squeeze theorem and the scalar rule for limits (The squeeze theorem, Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Absolute value and order: $|t| = 1$ forces $t = 1$ or $t = -1$ ; $0 < 1$ , so $1/2 < 1 < 2$ and $1/8 < 2$ , and $1/2 \ne 2$ ; reciprocals reverse the order; multiplying by a positive preserves it (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, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)).

[L8]

The true chain of inequalities relating the four quantities (For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ ).

[L9]

The refuted claim: for every sequence of positive reals, $\limsup_k a_{k+1}^{1/(k+1)} = \limsup_k a_{k+1}/a_k$ .

Refutation

technique · direct

1.1

Each $s_k$ is $1$ or $-1$ because $|s_k| = 1$ , so $t_k$ is well defined, with $t_k \in \{2, 1/2\}$ and $t_k > 0$ ; hence $a_k = 2^{-k} t_k > 0$ for every $k$ , and $(a_k)$ is a sequence of positive reals to which the claim applies. This is the sequence usually written $a_k = 2^{-k + (-1)^k}$ .

givenL1L4L7L9

1.2

Since $s_{k+1} = -s_k$ and $1 \ne -1$ , exactly one of the two situations " $s_k = 1$ and $s_{k+1} = -1$ " and " $s_k = -1$ and $s_{k+1} = 1$ " occurs at each index $k$ . In the first, $t_k = 2$ and $t_{k+1} = 1/2$ ; in the second, $t_k = 1/2$ and $t_{k+1} = 2$ .

givenL1L7

1.3

For every $n$ both values of $s$ occur at an index $\ge n$ : $s_{e_n} = 1$ with $e_n \ge n$ and $s_{o_n} = -1$ with $o_n \ge n$ .

givenL1

2.1

The ratios are $q_k = a_{k+1}/a_k = \big(2^{-(k+1)} t_{k+1}\big)/\big(2^{-k} t_k\big) = 2^{-1} t_{k+1}/t_k$ , which by step 1.2 equals $2^{-1}(1/2)/2 = 1/8$ when $s_k = 1$ and $2^{-1} \cdot 2/(1/2) = 2$ when $s_k = -1$ .

step 1.1step 1.2L4L7algebra

2.2

The roots are $r_k = \big(2^{-(k+1)} t_{k+1}\big)^{1/(k+1)} = \big(2^{-(k+1)}\big)^{1/(k+1)} t_{k+1}^{1/(k+1)} = 2^{-1} t_{k+1}^{1/(k+1)}$ .

step 1.1L4

2.3

Since $1/2 \le t_{k+1} \le 2$ for every $k$ and $x \mapsto x^{1/(k+1)}$ is nondecreasing on the positive reals, $(1/2)^{1/(k+1)} \le t_{k+1}^{1/(k+1)} \le 2^{1/(k+1)}$ ; both bounding sequences converge to $1$ by [L5], so the squeeze theorem gives $t_{k+1}^{1/(k+1)} \to 1$ .

step 1.1L4L5L6L7

3.1

By steps 1.2 and 1.3 the tail range of $(q_k)$ at every index $n$ is exactly $\{1/8, 2\}$ : those are the only values, and each occurs at some index $\ge n$ . Its least upper bound in $\overline{\mathbb{R}}$ is $2$ and its greatest lower bound is $1/8$ , since $1/8 < 2$ and both belong to the set; hence $\limsup_k q_k$ is the greatest lower bound of $\{2\}$ , namely $2$ , and $\liminf_k q_k$ is the least upper bound of $\{1/8\}$ , namely $1/8$ .

step 2.1step 1.3L2L7

3.2

By steps 2.2 and 2.3 and the scalar rule, $r_k = 2^{-1} t_{k+1}^{1/(k+1)} \to 2^{-1} \cdot 1 = 1/2$ , so $\limsup_k r_k = \liminf_k r_k = 1/2$ .

step 2.2step 2.3L3L6

4.1

For this sequence the claim asserts $\limsup_k r_k = \limsup_k q_k$ , that is $1/2 = 2$ ; but $1/2 < 1 < 2$ , so the two are different and the claim fails.

step 3.1step 3.2L7L9

5.1

The claim is therefore false. The true chain [L8] reads here $1/8 \le 1/2 \le 1/2 \le 2$ , so both outer inequalities are strict for this witness while the middle one is an equality.

step 4.1step 3.1step 3.2L7L8L9∎

Remarks

The witness is named on the companion page as $a_k = 2^{-k + (-1)^k}$ has $\liminf a_{k+1}/a_k = 1/8$ , $\limsup a_{k+1}/a_k = 2$ and $\lim a_k^{1/k} = 1/2$ ↗, which quotes the four values computed here.
This is the standard witness that the root criterion is strictly stronger. The ratios oscillate between $1/8$ and $2$ , so a criterion reading only $\limsup_k q_k = 2 > 1$ learns nothing about whether $a_k \to 0$ ; the roots converge to $1/2 < 1$ , which settles it. The same sequence reappears wherever the ratio and root tests are compared.
Why the roots are so much better behaved. Taking an $n$ -th root divides the exponent by $n$ , so the bounded perturbation $(-1)^k$ in the exponent of $2^{-k+(-1)^k}$ contributes $2^{\pm 1/(k+1)}$ , which tends to $1$ . The ratio, by contrast, differences the exponent, and a bounded oscillation does not shrink under differencing.
Both outer inequalities of For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ are strict here, but the middle one is not. A witness making all three strict at once is A positive sequence making all three inequalities of the ratio-to-root chain strict ↗.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 121 results over 35 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Root test (Wikipedia) (standard reference, not scraped)
Ratio test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.35, 3.37) (standard reference, not scraped)
N. Donaldson, Math 140A: Real Analysis notes (standard reference, not scraped)

FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak\limsup a_k^{1/k} = \limsup a_{k+1}/a_klimsupak1/k​=limsupak+1​/ak​ for every positive sequence