CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $\mathbb{N}$ 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

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

Its ratio and root families, in the shifted form used throughout, $q_k = |a_{k+1}/a_k|$ and $\rho_k = |a_{k+1}|^{1/(k+1)}$ , satisfy

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

as computed in FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ for every positive sequence. So the root test gives convergence of $\sum_{k \ge 1} |a_k|$ , while neither half of the ratio test applies: its convergence half needs $\limsup_k q_k < 1$ and $2$ is not below $1$ , and its divergence half needs $\liminf_k q_k > 1$ 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 $\mathbb{N}$ 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 $s \circ o \equiv -1$ ; $t_k := 2$ when $s_k = 1$ and $t_k := 1/2$ when $s_k = -1$ ; $a_k := 2^{-k} t_k$ ; and the families $q_k = |a_{k+1}/a_k|$ , $\rho_k = |a_{k+1}|^{1/(k+1)}$ (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 $\overline{\mathbb{R}}$ , 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 $\mathbb{N}$ 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 $s \circ o \equiv -1$ , Basic properties of the absolute value).

[L2]

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

[L3]

For this sequence, $\liminf_k q_k = 1/8$ , $\limsup_k q_k = 2$ and the root family converges to $1/2$ , so $\limsup_k \rho_k = 1/2$ (FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ for every positive sequence).

[L4]

The root test: $\limsup_k \rho_k < 1$ gives convergence of $\sum_{k \ge 1} |a_k|$ (Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L5]

The ratio test: its convergence half needs $\limsup_k q_k < 1$ and its divergence half needs $\liminf_k q_k > 1$ ; those are its only two criteria (Ratio test: $\limsup |a_{k+1}/a_k| < 1$ gives absolute convergence and hence convergence, and $\liminf |a_{k+1}/a_k| > 1$ gives divergence).

[L6]

Limit superior and inferior exist in $\overline{\mathbb{R}}$ for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, 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 $\overline{\mathbb{R}}$ ); 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 $a_k \ne 0$ , so both the ratio and the root families are defined and $|a_k| = a_k$ .

givenL1L2

2.1

For this sequence $\limsup_k q_k = 2$ and $\liminf_k q_k = 1/8$ .

step 1.1L3

2.2

For this sequence $\limsup_k \rho_k = 1/2$ .

step 1.1L3

3.1

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

step 2.2step 1.1L4L6

3.2

The convergence half of the ratio test does not apply, since $\limsup_k q_k = 2$ and $2 < 1$ is false.

step 2.1L5

3.3

The divergence half does not apply either, since $\liminf_k q_k = 1/8$ 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: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ 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)ka_k = 2^{-k+(-1)^k}ak​=2−k+(−1)k has ratio limsup 222 and liminf 1/81/81/8, so the ratio test fails, while the root test gives convergence