CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$a_{k} = 2^{- k + (- 1)^{k}}$ has ratio limsup $2$ and liminf $1 / 8$ , so the ratio test fails, while the root test gives convergence

Statement refuted

Refuted claim: whenever the root test decides a series, the ratio test decides it too; equivalently, the ratio test is no weaker than the root test.

The claim is refuted by the sequence usually written $a_{k} = 2^{- k + (- 1)^{k}}$ . Precisely, 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

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

Its ratio and root families, in the shifted form used throughout, $q_{k} = ∣ a_{k + 1} / a_{k} ∣$ and $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , satisfy

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

as computed in FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence. So the root test gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ , while neither half of the ratio test applies: its convergence half needs ${lim sup}_{k} q_{k} < 1$ and $2$ is not below $1$ , and its divergence half needs ${lim inf}_{k} q_{k} > 1$ and $1 / 8$ is not above $1$ .

This is the concrete form of the strict dominance recorded in Whenever the ratio test decides, the root test decides the same way, and the converse fails.

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 $s \circ o \equiv - 1$ ; $t_{k} : = 2$ when $s_{k} = 1$ and $t_{k} : = 1 / 2$ when $s_{k} = - 1$ ; $a_{k} : = 2^{- k} t_{k}$ ; and the families $q_{k} = ∣ a_{k + 1} / a_{k} ∣$ , $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ (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 $\overline{R}$ , Rational powers $a^{r}$ of a positive base).

[L1]

The alternating sequence satisfies $∣ s_{k} ∣ = 1$ for every $k$ , so each $s_{k}$ is $1$ or $- 1$ and $t_{k}$ is well defined with $t_{k} > 0$ (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 $s \circ o \equiv - 1$ , Basic properties of the absolute value).

[L2]

$2^{- k} > 0$ for every $k \in N$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Laws of integer exponents).

[L3]

For this sequence, ${lim inf}_{k} q_{k} = 1 / 8$ , ${lim sup}_{k} q_{k} = 2$ and the root family converges to $1 / 2$ , so ${lim sup}_{k} ρ_{k} = 1 / 2$ (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence).

[L4]

The root test: ${lim sup}_{k} ρ_{k} < 1$ gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L5]

The ratio test: its convergence half needs ${lim sup}_{k} q_{k} < 1$ and its divergence half needs ${lim inf}_{k} q_{k} > 1$ ; those are its only two criteria (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

[L6]

Limit superior and inferior exist in $\overline{R}$ for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, 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 $\overline{R}$ ); a series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series).

Counterexample

technique · direct

1.1

Each $s_{k}$ is $1$ or $- 1$ , so $t_{k}$ is defined and positive, and $a_{k} = 2^{- k} t_{k} > 0$ ; in particular $a_{k} \neq 0$ , so both the ratio and the root families are defined and $∣ a_{k} ∣ = a_{k}$ .

givenL1L2

2.1

For this sequence ${lim sup}_{k} q_{k} = 2$ and ${lim inf}_{k} q_{k} = 1 / 8$ .

step 1.1L3

2.2

For this sequence ${lim sup}_{k} ρ_{k} = 1 / 2$ .

step 1.1L3

3.1

Since $1 / 2 < 1$ , the root test applies and gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ , hence of $\sum_{k \geq 1} a_{k}$ , the terms being positive.

step 2.2step 1.1L4L6

3.2

The convergence half of the ratio test does not apply, since ${lim sup}_{k} q_{k} = 2$ and $2 < 1$ is false.

step 2.1L5

3.3

The divergence half does not apply either, since ${lim inf}_{k} q_{k} = 1 / 8$ and $1 / 8 > 1$ is false.

step 2.1L5

4.1

So the root test decides this series and the ratio test decides nothing about it, refuting the claim.

step 3.1step 3.2step 3.3∎

Remarks

The two families are computed once, on the previous page, and cited here. The four limit quantities for this sequence are established in the refutation of FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence, where the same witness shows that the outer inequalities of the ratio-to-root chain are strict. Nothing is recomputed here; what is added is the reading of those numbers through the two tests.
Why the roots behave and the ratios do not. The exponent of $a_{k}$ is $- k + (- 1)^{k}$ . Taking an $(k + 1)$ -st root divides that exponent by $k + 1$ , so the bounded oscillation contributes $2^{\pm 1 / (k + 1)}$ , which tends to $1$ ; forming a ratio differences the exponent, and a bounded oscillation does not shrink under differencing but doubles.
This does not make the root test universal. The companion counterexample with root limit exactly $1$ shows the root test has its own blind spot, and FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test shows no test on this page can avoid having one.

Depends on

Used by

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Sources

Root test (Wikipedia) (standard reference, not scraped)
Ratio test (Wikipedia) (standard reference, not scraped)
CSUDH notes on the ratio and root tests (standard reference, not scraped)

ak=2−k+(−1)kak​=2−k+(−1)k has ratio limsup 22 and liminf 1/81/8, so the ratio test fails, while the root test gives convergence