TheoremStatement: AI-adaptedProof: 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.

Gauss: for positive terms, if $a_k/a_{k+1} = 1 + h/k + r_k$ with $|r_k| \le C\,k^{-1-\varepsilon}$ for $k \ge 1$ , some constant $C$ and some rational $\varepsilon > 0$ , the series converges iff $h > 1$

Statement

Let $(a_k)$ be a sequence of reals with $a_k > 0$ for every $k \in \mathbb{N}$ . Suppose there are a real $h$ , a real $C \ge 0$ , a rational $\varepsilon > 0$ and reals $r_k$ for $k \ge 1$ such that

$\frac{a_k}{a_{k+1}} \;=\; 1 + \frac{h}{k} + r_k \qquad \text{and} \qquad |r_k| \;\le\; C\,k^{-1-\varepsilon} \qquad (k \ge 1),$

where $k$ denotes the canonical natural $\iota(k) > 0$ and $k^{-1-\varepsilon}$ is the rational power (Rational powers $a^r$ of a positive base, Canonical naturals are positive and strictly increasing). Then

$\sum a_k \ \text{converges} \qquad \Longleftrightarrow \qquad h > 1 .$

The hypotheses are imposed from $k = 1$ on, since $h/k$ has no value at $k = 0$ ; $a_0$ is unconstrained beyond being positive, which costs nothing because convergence is a tail property (A series converges iff each of its tail series converges, and the sum splits as $s_N$ plus the $N$ -th tail).

The exponent $\varepsilon$ is rational because that is what Rational powers $a^r$ of a positive base supplies, and the error bound is a $p$ -series bound with $p = 1 + \varepsilon > 1$ , which is exactly the summability the proof consumes.

The borderline case $h = 1$ is the whole point of the theorem. There $R_k = (k+1)(a_k/a_{k+1} - 1)$ tends to $1$ , so both halves of Raabe's test (Raabe is Kummer with $\zeta_k = k+1$ : for positive terms, $\liminf\, (k+1)(a_k/a_{k+1} - 1) > 1$ gives convergence and $\limsup < 1$ gives divergence) are silent; the theorem asserts divergence there, and the argument below establishes it without any logarithm, by a telescoping product estimate.

Facts & Assumptions

Given: A sequence $(a_k)$ of reals with $a_k > 0$ for every $k$ ; reals $h$ , $C \ge 0$ , a rational $\varepsilon > 0$ and reals $r_k$ ( $k \ge 1$ ) with $a_k/a_{k+1} = 1 + h/k + r_k$ and $|r_k| \le C k^{-1-\varepsilon}$ for $k \ge 1$ ; and $R_k := (k+1)(a_k/a_{k+1} - 1)$ for $k \in \mathbb{N}$ (Raabe is Kummer with $\zeta_k = k+1$ : for positive terms, $\liminf\, (k+1)(a_k/a_{k+1} - 1) > 1$ gives convergence and $\limsup < 1$ gives divergence).

[L1]

Raabe's test: for positive terms, $\liminf_k R_k > 1$ gives convergence of $\sum a_k$ and $\limsup_k R_k < 1$ gives divergence (Raabe is Kummer with $\zeta_k = k+1$ : for positive terms, $\liminf\, (k+1)(a_k/a_{k+1} - 1) > 1$ gives convergence and $\limsup < 1$ gives divergence).

[L2]

Every subset of $\overline{\mathbb{R}}$ has least upper and greatest lower bounds there; $\liminf_k x_k = \sup\{i_n\}$ and $\limsup_k x_k = \inf\{s_n\}$ for the tail bounds, both existing for every sequence (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, 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 tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence).

[L3]

For every real $c > 0$ there is a natural $n \ge 1$ with $1/n < c$ , and for every real $x$ there is a natural $n$ with $\iota(n) > x$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean).

[L4]

Rational powers on a positive base: $a^{r+s} = a^{r}a^{s}$ , $(a^{r})^{s} = a^{rs}$ , $a^{-r} = 1/a^{r}$ , $a^{r} > 0$ ; and for rational $t > 0$ , $0 < x < y$ implies $x^{t} < y^{t}$ (Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Rational powers $a^r$ of a positive base).

[L5]

Limit rules: sums, scalar multiples, products and quotients of convergent sequences (Algebra of limits: sums, scalar multiples, products and quotients); the squeeze theorem (The squeeze theorem); convergence depends only on the tail (Convergence depends only on the tail); a convergent sequence satisfies its estimate for every real tolerance (Limits and Cauchy sequences of reals, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L6]

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

[L7]

Direct comparison, and the fact that scaling by a nonzero constant preserves convergence and divergence (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 $\sum b_k$ , Convergent series add and scale termwise); a series converges if and only if each of its tail series converges, and for nonnegative terms every partial sum is at most the sum (A series converges iff each of its tail series converges, and the sum splits as $s_N$ plus the $N$ -th tail, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L8]

The principle of induction (The principle of mathematical induction); reciprocation reverses order on the positives (Inverses of positives are positive, and reciprocation reverses order); $|x| \ge 0$ and $x \le |x|$ (Basic properties of the absolute value).

Proof

technique · cases

1.1

Assume $h > 1$ .

assume-case gt

1.2

Assume instead $h < 1$ .

assume-case lt

1.3

Assume instead $h = 1$ .

assume-case eq

1.4

For every $k \ge 1$ , $R_k = (k+1)\left(\dfrac{h}{k} + r_k\right) = h\left(1 + \dfrac{1}{k}\right) + (k+1) r_k$ .

givenalgebra

1.5

The sequence $j \mapsto 1/(j+1)$ converges to $0$ : given a rational $\eta > 0$ , choose a natural $n \ge 1$ with $1/n < \eta$ ; then $1/(j+1) \le 1/n < \eta$ for every $j$ with $j+1 \ge n$ .

L3L8choose

1.6

The sequence $j \mapsto (j+1)^{-\varepsilon}$ converges to $0$ : given a real $\eta > 0$ , put $M := \max\{1/\eta,\, 1\} > 0$ and choose a natural $n$ with $\iota(n) > M^{1/\varepsilon}$ ; then for $j + 1 \ge n$ we get $(j+1)^{\varepsilon} > \big(M^{1/\varepsilon}\big)^{\varepsilon} = M \ge 1/\eta$ , hence $0 < (j+1)^{-\varepsilon} < \eta$ .

L3L4L8choose

1.7

For every $j \in \mathbb{N}$ , $|(j+2)\,r_{j+1}| \le C (j+2)(j+1)^{-1-\varepsilon} \le 2C (j+1)\,(j+1)^{-1-\varepsilon} = 2C\,(j+1)^{-\varepsilon}$ , using $j + 2 \le 2(j+1)$ .

givenL4L8algebra

2.1

Hence $-2C(j+1)^{-\varepsilon} \le (j+2) r_{j+1} \le 2C(j+1)^{-\varepsilon}$ with both bounds converging to $0$ , so $(j+2) r_{j+1} \to 0$ by the squeeze theorem.

step 1.7step 1.6L5

2.2

In the case $h = 1$ , put $u_k := \dfrac{k}{k+1}\,r_k$ and $t_k := k\,a_k$ for $k \ge 1$ ; then $t_k > 0$ , $|u_k| \le |r_k| \le C k^{-1-\varepsilon}$ since $0 < k/(k+1) < 1$ , and $\dfrac{a_k}{a_{k+1}} = \dfrac{k+1}{k} + r_k = \dfrac{k+1}{k}\big(1 + u_k\big)$ .

step 1.3givenL8algebra

3.1

Therefore $R_{j+1} = h\big(1 + 1/(j+1)\big) + (j+2) r_{j+1} \to h(1+0) + 0 = h$ , and since convergence depends only on the tail, the sequence $(R_k)_{k \in \mathbb{N}}$ converges to $h$ .

step 1.4step 1.5step 2.1L5

3.2

Consequently $\dfrac{t_k}{t_{k+1}} = \dfrac{k}{k+1}\cdot\dfrac{a_k}{a_{k+1}} = 1 + u_k$ for $k \ge 1$ , so $1 + u_k = t_k/t_{k+1} > 0$ and $\dfrac{t_{k+1}}{t_k} = \dfrac{1}{1+u_k}$ .

step 2.2algebra

3.3

In the case $h = 1$ : $\sum_{k \ge 1} k^{-1-\varepsilon}$ converges, since $1 + \varepsilon$ is a rational exceeding $1$ ; hence so does $\sum_{k \ge 1} C k^{-1-\varepsilon}$ , and by comparison with it so does $\sum_{k \ge 1} |u_k|$ , whose terms are nonnegative.

step 2.2L6L7

4.1

In the case $h > 1$ : applying the limit estimate with the real tolerance $(h-1)/2 > 0$ gives an $N$ with $R_k > h - (h-1)/2 = (h+1)/2$ for all $k \ge N$ ; so $(h+1)/2$ is a lower bound of $\{R_k : k \ge N\}$ , whence $\liminf_k R_k \ge (h+1)/2 > 1$ and $\sum a_k$ converges.

step 1.1step 3.1L2L5L1

4.2

In the case $h < 1$ : the tolerance $(1-h)/2 > 0$ gives an $N$ with $R_k < h + (1-h)/2 = (h+1)/2$ for all $k \ge N$ ; so $(h+1)/2$ is an upper bound of $\{R_k : k \ge N\}$ , whence $\limsup_k R_k \le (h+1)/2 < 1$ and $\sum a_k$ diverges.

step 1.2step 3.1L2L5L1

4.3

For $k \ge 1$ : $(1-u_k)(1+u_k) = 1 - u_k^{2} \le 1$ and $1 + u_k > 0$ , so $\dfrac{1}{1+u_k} \ge 1 - u_k \ge 1 - |u_k|$ .

step 3.2L8algebra

4.4

Writing $U$ for the sum of $\sum_{k \ge 1} |u_k|$ and $P_n = \sum_{k=1}^{n} |u_k|$ for its partial sums, $P_n \to U$ , so there is a natural $N_0$ with $U - P_{N_0} \le 1/2$ ; put $N := N_0 + 1$ .

step 3.3L5L7choose

5.1

For every $n \ge N$ the block $\sum_{k=N}^{n} |u_k|$ is a partial sum of the $N_0$ -th tail series of $\sum_{k \ge 1}|u_k|$ , whose terms are nonnegative and whose sum is $U - P_{N_0}$ ; hence $\sum_{k=N}^{n} |u_k| \le 1/2$ , and in particular $|u_k| \le 1/2$ for every $k \ge N$ .

step 4.4L7

6.1

In the case $h = 1$ : for every $n \ge N-1$ , $\dfrac{t_{n+1}}{t_N} \ge 1 - \sum_{k=N}^{n} |u_k|$ , by induction on $n$ . At $n = N-1$ both sides equal $1$ , the sum being empty. Assume it at $n$ ; then, since $t_{n+1}/t_N > 0$ and $1/(1+u_{n+1}) \ge 1 - |u_{n+1}| \ge 1/2 > 0$ , and since the induction hypothesis gives $t_{n+1}/t_N \ge 1 - \sum_{k=N}^{n}|u_k| \ge 1/2 > 0$ , we get $\dfrac{t_{n+2}}{t_N} = \dfrac{1}{1+u_{n+1}}\cdot\dfrac{t_{n+1}}{t_N} \ge \big(1-|u_{n+1}|\big)\Big(1 - \sum_{k=N}^{n}|u_k|\Big) \ge 1 - \sum_{k=N}^{n+1}|u_k|$ , the last step expanding the product and discarding a nonnegative term.

step 3.2step 4.3step 5.1L8

7.1

Hence $t_{n+1}/t_N \ge 1 - 1/2 = 1/2$ for every $n \ge N-1$ , that is $m\,a_m = t_m \ge t_N/2 > 0$ for every $m \ge N$ , and so $a_m \ge \dfrac{t_N}{2}\cdot\dfrac{1}{m}$ for every $m \ge N$ .

step 6.1step 5.1L8algebra

8.1

The series $\sum_{m \ge 1} 1/m$ diverges, so $\sum_{m \ge 1} \frac{t_N}{2}\cdot\frac{1}{m}$ diverges, the factor $t_N/2$ being nonzero.

step 7.1L6L7

9.1

If $\sum a_k$ converged, then so would $\sum_{m \ge 1} a_m$ , and comparison with the estimate of step 7.1 would make $\sum_{m \ge 1} \frac{t_N}{2m}$ converge, contradicting step 8.1; so in the case $h = 1$ the series $\sum a_k$ diverges.

step 7.1step 8.1L7

10.1

The three cases $h > 1$ , $h < 1$ and $h = 1$ exhaust the reals, and they give convergence, divergence and divergence respectively; so $\sum a_k$ converges exactly when $h > 1$ .

step 4.1step 4.2step 9.1cases-exhaustive∎

Remarks

No logarithm anywhere, and that is deliberate. The classical treatment of $h = 1$ compares $a_k$ with $1/(k \log k)$ or invokes Bertrand's test. Neither is available in this library at this point, and neither is needed: the hypothesis $|r_k| \le C k^{-1-\varepsilon}$ makes $\sum |u_k|$ convergent, and a convergent sum of nonnegative errors is exactly what the product estimate of step 6.1 consumes. The price is that the theorem is stated with an $\varepsilon$ of decay to spare, rather than for an arbitrary summable error.
Step 7.1 is the Weierstrass product inequality in disguise. In the form $\prod_{j}(1 - x_j) \ge 1 - \sum_j x_j$ for $x_j \in [0,1]$ , it is the standard statement; here the product is $t_{n+1}/t_N$ , telescoped in advance, so that one induction does the work of two and no separate lemma about products of inequalities is needed.
What the conclusion at $h = 1$ says about the terms. The estimate $a_m \ge (t_N/2)\,(1/m)$ is a genuine lower bound of harmonic type: at the borderline the terms cannot decay faster than a constant multiple of $1/m$ , and divergence follows from the divergence of the harmonic series alone.
The three cases are decided by $h$ and by nothing else. The constants $C$ and $\varepsilon$ never appear in the conclusion; they enter only through the requirement that the error be summable, which is what keeps the case $h = 1$ from being genuinely borderline in this argument.

Depends on

Used by

How the nonnegative tests are ordered by strength, and which of them this page cannot state without the logarithm Remark

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 123 results over 33 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

Convergence tests (Wikipedia) (standard reference, not scraped)
K. Knopp, Theory and Application of Infinite Series, Ch. IX (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)