ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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 series with ratio limit exactly $1$ that Raabe decides

Example

Take $a_{k} : = 1 / ι (k + 1)^{2}$ for $k \in N$ , so that $\sum a_{k}$ is $\sum_{k \geq 1} 1 / k^{2}$ (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

its ratios $q_{k} = a_{k + 1} / a_{k} = \frac{(k + 1)^{2}}{(k + 2)^{2}}$ converge to $1$ , so neither half of the ratio test applies (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);
its Raabe expression is exactly $R_{k} = (k + 1) (\frac{a_{k}}{a_{k + 1}} - 1) = \frac{(k + 2)^{2} - (k + 1)^{2}}{k + 1} = \frac{2 k + 3}{k + 1} = 2 + \frac{1}{k + 1},$ which exceeds $2$ at every index, so ${lim inf}_{k} R_{k} \geq 2 > 1$ and Raabe's test gives convergence (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence).

This is the smallest honest illustration that Raabe's test decides series the ratio test cannot. The verdict agrees with For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ at $p = 2$ , as it must.

Facts & Assumptions

Given: The sequence $a_{k} : = 1 / ι (k + 1)^{2}$ , $k \in N$ ; its ratios $q_{k} = a_{k + 1} / a_{k}$ ; and its Raabe expression $R_{k} = (k + 1) (a_{k} / a_{k + 1} - 1)$ (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence, Integer powers $a^{m}$ , Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals are positive, so every $a_{k}$ is positive; reciprocation on the positives is order reversing; and $x^{2} = x \cdot x$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L2]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ , so $1 / ι (k + 1) \to 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Limits and Cauchy sequences of reals).

[L3]

Algebra of limits: sums, products and quotients of convergent sequences converge, the quotient requiring a nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients).

[L4]

The ratio test: its convergence half needs ${lim sup}_{k} q_{k} < 1$ and its divergence half needs ${lim inf}_{k} q_{k} > 1$ (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).

[L5]

Raabe's test: ${lim inf}_{k} R_{k} > 1$ gives convergence (Raabe is Kummer with $ζ_{k} = k + 1$ : for positive terms, $lim inf (k + 1) (a_{k} / a_{k + 1} - 1) > 1$ gives convergence and $lim sup < 1$ gives divergence).

[L6]

Limit superior and inferior in $\overline{R}$ , their existence for every sequence, and the descriptions ${lim sup}_{k} x_{k} = \inf {s_{n}}$ , ${lim inf}_{k} x_{k} = \sup {i_{n}}$ of the tail bounds (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 tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, 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).

[L7]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Verification

technique · direct

1.1

Every $a_{k} = 1 / ι (k + 1)^{2}$ is positive, so the ratios and the Raabe expression are defined.

givenL1

2.1

The ratios are $q_{k} = \frac{1 / ι (k + 2)^{2}}{1 / ι (k + 1)^{2}} = \frac{ι (k + 1)^{2}}{ι (k + 2)^{2}} = {(1 - \frac{1}{ι (k + 2)})}^{2}$ .

step 1.1L1algebra

2.2

The Raabe expression is $R_{k} = (k + 1) (\frac{ι (k + 2)^{2}}{ι (k + 1)^{2}} - 1) = \frac{ι (k + 2)^{2} - ι (k + 1)^{2}}{ι (k + 1)} = \frac{2 ι (k) + 3}{ι (k) + 1} = 2 + \frac{1}{ι (k + 1)}$ .

step 1.1L1algebra

3.1

Since $1 / ι (k + 2) \to 0$ , the product rule gives $q_{k} \to (1 - 0)^{2} = 1$ .

step 2.1L2L3

3.2

From step 2.2, $R_{k} > 2$ for every $k \in N$ , the added term $1 / ι (k + 1)$ being positive.

step 2.2L1

4.1

The convergence half of the ratio test does not apply: if ${lim sup}_{k} q_{k} < 1$ , then with $t$ real and ${lim sup}_{k} q_{k} < t < 1$ some tail supremum would be below $t$ , putting $q_{k} \leq t < 1$ for all large $k$ and contradicting $q_{k} \to 1$ .

step 3.1L4L6

4.2

The divergence half does not apply either: if ${lim inf}_{k} q_{k} > 1$ , some tail infimum would exceed $1$ , putting $q_{k} \geq c > 1$ for all large $k$ and again contradicting $q_{k} \to 1$ .

step 3.1L4L6

4.3

On the other hand $2$ is a lower bound of ${R_{k} : k \geq 0}$ , so the tail infimum $i_{0} \geq 2$ and ${lim inf}_{k} R_{k} \geq 2 > 1$ .

step 3.2L6

5.1

Raabe's test therefore gives convergence of $\sum a_{k}$ , that is of $\sum_{k \geq 1} 1 / k^{2}$ , in agreement with the case $p = 2$ of the $p$ -series theorem.

step 4.3step 1.1L5L7∎

Remarks

The Raabe expression here is exact, not asymptotic. Step 2.2 computes $R_{k} = 2 + 1 / (k + 1)$ on the nose, so no limit is needed to apply the test: a single inequality $R_{k} > 2$ at every index already forces ${lim inf}_{k} R_{k} \geq 2$ . That is why this witness is the cleanest available one.
Why the ratio test must fail here. The ratios of any $p$ -series tend to $1$ whatever $p$ is, so a criterion reading only $lim sup$ and $lim inf$ of the ratios cannot separate the convergent $p$ -series from the divergent ones. Raabe reads the rate at which the ratios approach $1$ , which is exactly the missing information, and that rate is $2 / k$ up to smaller terms when $p = 2$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

66 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Raabe's test (Wikipedia) (standard reference, not scraped)
Ratio test (Wikipedia) (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)
Binghamton University notes on Kummer, Raabe, and Gauss tests (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

A series with ratio limit exactly 11 that Raabe decides