False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: $\limsup |a_{k+1}/a_k| \ge 1$ implies the series diverges

Statement

False claim: for every sequence $(a_k)$ of reals with $a_k \ne 0$ for every $k$ , if

$\limsup_{k} \Big|\frac{a_{k+1}}{a_k}\Big| \;\ge\; 1$

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 $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ ).

The true divergence half of the ratio test (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) has the hypothesis $\liminf_k |a_{k+1}/a_k| > 1$ , 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 $\mathbb{N}$ 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 $s \circ o \equiv -1$ , put $c_k := 2 + s_k$ , so $c_k$ is $3$ at the even indices and $1$ at the odd ones, and let

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

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 $\mathbb{N}$ 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 $s \circ o \equiv -1$ ; $c_k := 2 + s_k$ ; and $a_k := c_k 2^{-k}$ for $k \in \mathbb{N}$ .

[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 $\mathbb{N}$ 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]

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 $\overline{\mathbb{R}}$ : $\limsup_k x_k = \inf\{s_n\}$ with $s_n = \sup\{x_k : k \ge n\}$ ; every subset of $\overline{\mathbb{R}}$ 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 $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$ , 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 $\mathbb{R}$ , The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined, The tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , 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 $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges, Convergent series add and scale termwise, If $0 \le a_k \le b_k$ 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: $\liminf_k |a_{k+1}/a_k| > 1$ gives divergence, and that is the only divergence criterion it supplies (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]

The refuted claim: for every sequence of nonzero reals with $\limsup_k |a_{k+1}/a_k| \ge 1$ , the series $\sum a_k$ diverges.

Refutation

technique · direct

1.1

Each $s_k$ is $1$ or $-1$ , since $|s_k| = 1$ ; so $c_k = 2 + s_k$ is $3$ or $1$ , in either case $1 \le c_k \le 3$ and $c_k > 0$ .

givenL1

2.1

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

step 1.1L2L6

2.2

For every $k$ , $0 < a_k = c_k 2^{-k} \le 3 \cdot 2^{-k}$ .

step 1.1L2

3.1

The ratios are $q_k := |a_{k+1}/a_k| = a_{k+1}/a_k = \dfrac{c_{k+1}}{c_k}\cdot\dfrac{1}{2}$ , 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 $s_{k+1} = -s_k$ .

step 1.1step 2.1L1L2algebra

3.2

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 $\sum_k a_k$ .

step 2.2L2L4

4.1

For every $n \in \mathbb{N}$ there is an index $k \ge n$ with $s_k = -1$ , namely $k = o_n$ , since $o_n \ge n$ ; so $q_k = 3/2$ at some index $k \ge n$ , for every $n$ .

step 3.1L1

5.1

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

step 4.1L3

6.1

So $(a_k)$ has nonzero terms and $\limsup_k |a_{k+1}/a_k| \ge 1$ , yet $\sum a_k$ converges; the claim fails for it and is therefore false.

step 2.1step 5.1step 3.2L6

7.1

Nothing in the ratio test is contradicted: its divergence half requires $\liminf_k q_k > 1$ , and here $\liminf_k q_k \le 1/6 < 1$ , since $q_k = 1/6$ at indices $e_n \ge n$ for every $n$ .

step 6.1step 3.1L1L3L5∎

Remarks

Replacing $\liminf$ by $\limsup$ is already fatal, even with the inequality kept strict. The witness above has $\limsup_k q_k \ge 3/2$ , which is strictly greater than $1$ , and its series converges. So the false claim is not rescued by demanding $\limsup_k |a_{k+1}/a_k| > 1$ : the two quantities $\liminf$ and $\limsup$ 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 $\liminf$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 110 results over 28 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

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

FALSE: lim sup⁡∣ak+1/ak∣≥1\limsup |a_{k+1}/a_k| \ge 1limsup∣ak+1​/ak​∣≥1 implies the series diverges