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

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FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence

Statement

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

${lim sup}_{k} a_{k + 1}^{1 / (k + 1)} = {lim sup}_{k} \frac{a_{k + 1}}{a_{k}},$

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 $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ , since $a_{k}^{1 / k}$ is undefined at $k = 0$ ; classically the claim reads ${lim sup}_{n} a_{n}^{1 / n} = {lim sup}_{n} a_{n + 1} / a_{n}$ .)

What is true is the chain ${lim inf}_{k} \frac{a_{k + 1}}{a_{k}} \leq {lim inf}_{k} a_{k + 1}^{1 / (k + 1)} \leq {lim sup}_{k} a_{k + 1}^{1 / (k + 1)} \leq {lim sup}_{k} \frac{a_{k + 1}}{a_{k}}$ of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ . 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 ${lim inf}_{k} \frac{a_{k + 1}}{a_{k}} = \frac{1}{8}, {lim sup}_{k} \frac{a_{k + 1}}{a_{k}} = 2, \lim_{k} a_{k + 1}^{1 / (k + 1)} = \frac{1}{2},$ 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 $N$ 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 $t_{k}$ 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 $N$ 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 $n_{j} \geq j$ (A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L2]

Limit superior and limit inferior in $\overline{R}$ , 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 $\inf_{n} \sup_{k \geq n} x_{k}$ 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).

[L3]

A sequence converging to a real $c$ has $lim sup = lim inf = c$ (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 $\pm \infty$ ).

[L4]

Powers and roots of positive reals: integer powers with $2^{m} 2^{m^{'}} = 2^{m + m^{'}}$ and $2^{- m} = 1 / 2^{m}$ ; $(x y)^{1 / n} = x^{1 / n} y^{1 / n}$ ; the integer power is the rational power at an integer exponent, so $(2^{- n})^{1 / n} = 2^{- n / n} = 2^{- 1}$ ; roots of positive reals are positive; and $0 < x \leq y$ implies $x^{1 / n} \leq y^{1 / n}$ (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 $a^{1 / n} \geq 0$ 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 $1 / 2 \neq 2$ ; 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 $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ ).

[L9]

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

Refutation

technique · direct

1.1givenL1L4L7L9

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

1.2givenL1L7

Since $s_{k + 1} = - s_{k}$ and $1 \neq - 1$ , 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 $t_{k + 1} = 2$ .

1.3givenL1

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

2.1step 1.1step 1.2L4L7algebra

The ratios are $q_{k} = a_{k + 1} / a_{k} = (2^{- (k + 1)} t_{k + 1}) / (2^{- k} t_{k}) = 2^{- 1} t_{k + 1} / t_{k}$ , 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 $s_{k} = - 1$ .

2.2step 1.1L4

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

2.3step 1.1L4L5L6L7

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

3.1step 2.1step 1.3L2L7

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 $\geq n$ . Its least upper bound in $\overline{R}$ is $2$ and its greatest lower bound is $1 / 8$ , since $1 / 8 < 2$ and both belong to the set; hence ${lim sup}_{k} q_{k}$ is the greatest lower bound of ${2}$ , namely $2$ , and ${lim inf}_{k} q_{k}$ is the least upper bound of ${1 / 8}$ , namely $1 / 8$ .

3.2step 2.2step 2.3L3L6

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 ${lim sup}_{k} r_{k} = {lim inf}_{k} r_{k} = 1 / 2$ .

4.1step 3.1step 3.2L7L9

For this sequence the claim asserts ${lim sup}_{k} r_{k} = {lim sup}_{k} q_{k}$ , that is $1 / 2 = 2$ ; but $1 / 2 < 1 < 2$ , so the two are different and the claim fails.

5.1step 4.1step 3.1step 3.2L7L8L9∎

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

Remarks

The witness is named on the companion page as $a_{k} = 2^{- k + (- 1)^{k}}$ has $lim inf a_{k + 1} / a_{k} = 1 / 8$ , $lim sup a_{k + 1} / a_{k} = 2$ and $\lim a_{k}^{1 / k} = 1 / 2$ ↗, 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 ${lim sup}_{k} q_{k} = 2 > 1$ 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 $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ 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 ↗.

FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/aklimsupak1/k​=limsupak+1​/ak​ for every positive sequence

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence