False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges

Statement

False claim: for every sequence $(a_{k})$ of reals with $a_{k} \neq 0$ for every $k$ , if

${lim sup}_{k} ∣ \frac{a_{k + 1}}{a_{k}} ∣ \geq 1$

then $\sum a_{k}$ diverges (Series, partial sums, convergence and the sum, divergence, and the tail series, 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 true divergence half of the ratio test (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) has the hypothesis ${lim inf}_{k} ∣ a_{k + 1} / a_{k} ∣ > 1$ , on the limit inferior and with a strict inequality. The claim above weakens it in both respects at once, and either weakening alone already destroys it.

The witness is built from the alternating sequence: with $(s_{k})$ as in 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 $s \circ o \equiv - 1$ , put $c_{k} : = 2 + s_{k}$ , so $c_{k}$ is $3$ at the even indices and $1$ at the odd ones, and let

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

Its ratios take only the two values $3 / 2$ and $1 / 6$ , so their limit superior is at least $1$ , while the series converges by comparison with a geometric series.

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 $s \circ o \equiv - 1$ ; $c_{k} : = 2 + s_{k}$ ; and $a_{k} : = c_{k} 2^{- k}$ for $k \in N$ .

[L1]

The alternating sequence: $s_{0} = 1$ , $s_{k + 1} = - s_{k}$ , $∣ s_{k} ∣ = 1$ for every $k$ , $s_{e_{j}} = 1$ and $s_{o_{j}} = - 1$ , with $e$ and $o$ 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]

Powers of $2$ : $2^{- k} > 0$ , $2^{- (k + 1)} = 2^{- k} / 2$ , and $2^{- k} = 1 / 2^{k}$ (Integer powers $a^{m}$ , Laws of integer exponents, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L3]

Limit superior in $\overline{R}$ : ${lim sup}_{k} x_{k} = \inf {s_{n}}$ with $s_{n} = \sup {x_{k} : k \geq n}$ ; every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there; and both quantities exist 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 $\overline{R}$ , 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 $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L4]

The geometric series $\sum_{k} r^{k}$ converges for $∣ r ∣ < 1$ ; scaling by a nonzero constant preserves convergence; and direct comparison applies to nonnegative terms (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Convergent series add and scale termwise, If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , Basic properties of the absolute value).

[L5]

The ratio test: ${lim inf}_{k} ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence, and that is the only divergence criterion it supplies (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]

The refuted claim: for every sequence of nonzero reals with ${lim sup}_{k} ∣ a_{k + 1} / a_{k} ∣ \geq 1$ , the series $\sum a_{k}$ diverges.

Refutation

technique · direct

1.1givenL1

Each $s_{k}$ is $1$ or $- 1$ , since $∣ s_{k} ∣ = 1$ ; so $c_{k} = 2 + s_{k}$ is $3$ or $1$ , in either case $1 \leq c_{k} \leq 3$ and $c_{k} > 0$ .

2.1step 1.1L2L6

Hence $a_{k} = c_{k} 2^{- k} > 0$ for every $k$ , so in particular $a_{k} \neq 0$ and the claim applies to $(a_{k})$ .

2.2step 1.1L2

For every $k$ , $0 < a_{k} = c_{k} 2^{- k} \leq 3 \cdot 2^{- k}$ .

3.1step 1.1step 2.1L1L2algebra

The ratios are $q_{k} : = ∣ a_{k + 1} / a_{k} ∣ = a_{k + 1} / a_{k} = \frac{c_{k + 1}}{c_{k}} \cdot \frac{1}{2}$ , so $q_{k} = (1 / 3) (1 / 2) = 1 / 6$ when $s_{k} = 1$ and $q_{k} = 3 \cdot (1 / 2) = 3 / 2$ when $s_{k} = - 1$ , using $s_{k + 1} = - s_{k}$ .

3.2step 2.2L2L4

The geometric series $\sum_{k} (1 / 2)^{k}$ converges, since $∣ 1 / 2 ∣ < 1$ ; hence so does $\sum_{k} 3 \cdot 2^{- k}$ , and by comparison so does $\sum_{k} a_{k}$ .

4.1step 3.1L1

For every $n \in N$ there is an index $k \geq n$ with $s_{k} = - 1$ , namely $k = o_{n}$ , since $o_{n} \geq n$ ; so $q_{k} = 3 / 2$ at some index $k \geq n$ , for every $n$ .

5.1step 4.1L3

Therefore every tail supremum satisfies $\sup {q_{k} : k \geq n} \geq 3 / 2$ , so $3 / 2$ is a lower bound of the set of tail suprema and ${lim sup}_{k} q_{k} \geq 3 / 2 \geq 1$ .

6.1step 2.1step 5.1step 3.2L6

So $(a_{k})$ has nonzero terms and ${lim sup}_{k} ∣ a_{k + 1} / a_{k} ∣ \geq 1$ , yet $\sum a_{k}$ converges; the claim fails for it and is therefore false.

7.1step 6.1step 3.1L1L3L5∎

Nothing in the ratio test is contradicted: its divergence half requires ${lim inf}_{k} q_{k} > 1$ , and here ${lim inf}_{k} q_{k} \leq 1 / 6 < 1$ , since $q_{k} = 1 / 6$ at indices $e_{n} \geq n$ for every $n$ .

Remarks

Replacing $lim inf$ by $lim sup$ is already fatal, even with the inequality kept strict. The witness above has ${lim sup}_{k} q_{k} \geq 3 / 2$ , which is strictly greater than $1$ , and its series converges. So the false claim is not rescued by demanding ${lim sup}_{k} ∣ a_{k + 1} / a_{k} ∣ > 1$ : the two quantities $lim inf$ and $lim sup$ are genuinely different hypotheses here, and only the first one works.
The asymmetry of the ratio test is not an accident of its proof. A large ratio occurring arbitrarily late says only that the terms grow at those steps; it says nothing about their size, because they may have been made very small in between. Only an eventual lower bound on the ratios forces the terms to stay away from $0$ , and that is precisely a hypothesis on $lim inf$ .

FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: lim sup⁡∣ak+1/ak∣≥1limsup∣ak+1​/ak​∣≥1 implies the series diverges

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges