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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$a_k = (-1)^k$ , $b_k = k$ have $a_k/b_k \to 0$ while the difference quotient oscillates, so Stolz-Cesaro has no converse

Statement refuted

Refuted claim: the converse of Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ . That is: if $(b_k)$ is strictly increasing with range not bounded above and the quotients $a_k/b_k$ converge, then the difference quotients $(a_{k+1}-a_k)/(b_{k+1}-b_k)$ converge too.

The witness is $a_k = s_k$ , 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 $s \circ o \equiv -1$ usually written $(-1)^k$ , and $b_k = k$ . Then $b_0 = 0$ , so the quotient $a_k/b_k$ is formed for $k \ge 1$ only, exactly as Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ is stated; and there

$\frac{a_k}{b_k} = \frac{s_k}{k}, \qquad \Big|\frac{a_k}{b_k}\Big| = \frac1k \longrightarrow 0,$

while the difference quotient is

$\frac{a_{k+1}-a_k}{b_{k+1}-b_k} \;=\; s_{k+1} - s_k \;=\; -2 s_k,$

which takes the value $-2$ at even $k$ and $2$ at odd $k$ and does not converge.

Facts & Assumptions

Given: The alternating sequence $(s_k)$ with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ ; the sequences $a_k := s_k$ and $b_k := k \cdot 1_{\mathbb{R}}$ ; and the difference quotients $d_k := (a_{k+1}-a_k)(b_{k+1}-b_k)^{-1}$ .

[L1]

The alternating sequence: $|s_k| = 1$ for every $k$ , $s_{k+1} = -s_k$ , $s_{e_j} = 1$ and $s_{o_j} = -1$ along the even and odd index maps (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 $s \circ o \equiv -1$ ).

[L2]

$(s_k)$ is bounded and does not converge (FALSE: every bounded sequence converges).

[L3]

The canonical naturals of $\mathbb{R}$ are positive for $k \ge 1$ and strictly increasing (Canonical naturals are positive and strictly increasing), and no real bounds them all (Every complete ordered field is Archimedean); strict monotonicity and boundedness of real sequences (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Convergence of real sequences and uniqueness of limits (Limits and Cauchy sequences of reals, A sequence has at most one limit); convergence depends only on a tail (Convergence depends only on the tail); the reciprocal Archimedean property (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L5]

Algebra of limits, in particular that a scalar multiple of a convergent sequence converges (Algebra of limits: sums, scalar multiples, products and quotients).

[L6]

Order arithmetic: $|u| \ge 0$ and $|uv| = |u||v|$ (Basic properties of the absolute value); a positive element is invertible with positive inverse and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field).

[L7]

The hypotheses and conclusion of Stolz-Cesaro in the $\infty/\infty$ form (Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ ).

Counterexample

technique · direct

1.1

$(b_k)$ is strictly increasing, its range is not bounded above, $b_0 = 0$ and $b_k > 0$ for $k \ge 1$ ; so the hypotheses of Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ on $(b_k)$ hold with $K_0 = 1$ .

L3L7

1.2

$|s_k| = 1$ for every $k$ , and $(s_k)$ does not converge.

L1L2

2.1

The tail quotients $q_j := a_{j+1}/b_{j+1} = s_{j+1}(j+1)^{-1}$ satisfy $|q_j| = (j+1)^{-1}$ ; given a real $\varepsilon > 0$ and a natural $m \ge 1$ with $1/m < \varepsilon$ , every $j \ge m$ has $|q_j - 0| = (j+1)^{-1} < \varepsilon$ , so $q_j \to 0$ .

step 1.1step 1.2L4L6

2.2

$b_{k+1} - b_k = 1$ for every $k$ , so $d_k = a_{k+1} - a_k = s_{k+1} - s_k = -s_k - s_k = -2s_k$ , which is $-2$ when $k$ is even and $2$ when $k$ is odd.

step 1.1step 1.2L1L3

3.1

$(d_k)$ does not converge: were $d_k \to M$ , then $s_k = (-1/2)\,d_k$ would converge to $-M/2$ by the scalar-multiple rule, contradicting step 1.2.

step 1.2step 2.2L5

4.1

So $(b_k)$ is strictly increasing and unbounded, the quotients $a_k/b_k$ converge to $0$ over the indices $k \ge 1$ , and the difference quotients do not converge: the converse of Stolz-Cesaro is false.

step 2.1step 3.1L7∎

Remarks

The implication is genuinely one-way, and the reason is averaging. The conclusion of Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ is obtained by summing the difference quotients against the weights $b_{k+1}-b_k$ and dividing by $b_n$ , which is a weighted average. An average can converge while what is averaged oscillates, and here it does: the quotients $a_k/b_k$ are damped by the growing denominator, and no information about $d_k$ survives.
The same phenomenon, in the summability language. With $b_k = k$ the weights are equal, so this is the Cesaro situation of FALSE: if the Cesaro means of a sequence converge then the sequence converges and The Cesaro means of $(-1)^k$ converge to $0$ although the sequence diverges in another costume; the divergent object is the same alternating sequence.
The index range is not a technicality. $b_0 = 0$ , so $a_0/b_0$ does not denote anything, and the quotient sequence exists only from $k = 1$ . That is why Stolz-Cesaro, $\infty/\infty$ form: if $b_k$ is strictly increasing and unbounded and $(a_{k+1}-a_k)/(b_{k+1}-b_k) \to L$ then $a_k/b_k \to L$ is stated for the tail, and why the convergence asserted above is asserted over $k \ge 1$ .

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: 77 results over 24 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

Stolz-Cesàro theorem (Wikipedia) (standard reference, not scraped)
Limit of a sequence (Wikipedia) (standard reference, not scraped)

ak=(−1)ka_k = (-1)^kak​=(−1)k, bk=kb_k = kbk​=k have ak/bk→0a_k/b_k \to 0ak​/bk​→0 while the difference quotient oscillates, so Stolz-Cesaro has no converse