TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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.

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

Statement

Let $(a_{k})$ be a sequence of reals with $a_{k} \neq 0$ for every $k \in N$ and put

$q_{k} : = ∣ \frac{a_{k + 1}}{a_{k}} ∣ = \frac{∣ a_{k + 1} ∣}{∣ a_{k} ∣} (k \in N),$

a genuine sequence on $N$ , whose limit superior and limit inferior exist in $\overline{R}$ for every such $(a_{k})$ (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence). Then:

if ${lim sup}_{k} q_{k} < 1$ then $\sum ∣ a_{k} ∣$ converges, and hence $\sum a_{k}$ converges as well (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges);
if ${lim inf}_{k} q_{k} > 1$ then $\sum a_{k}$ diverges.

The hypothesis $a_{k} \neq 0$ is what makes $q_{k}$ exist and is not a convenience: a single vanishing term leaves the ratio at that index undefined. For a family from a starting index $m$ the statement is the one above applied to the shifted sequence $j \mapsto a_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series), whose ratios are $∣ a_{j + m + 1} / a_{j + m} ∣$ .

Nothing is claimed when ${lim sup}_{k} q_{k} \geq 1 \geq {lim inf}_{k} q_{k}$ . In that regime the test is silent, and it has to be: the companion page carries a convergent series whose ratios have limit superior $2$ , and both a convergent and a divergent series with ratio limit exactly $1$ .

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} \neq 0$ for every $k$ ; the ratios $q_{k} = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣$ ; the tail bounds $s_{n} = \sup {q_{k} : k \geq n}$ and $i_{n} = \inf {q_{k} : k \geq n}$ taken in $\overline{R}$ , so that ${lim sup}_{k} q_{k} = \inf {s_{n}}$ and ${lim inf}_{k} q_{k} = \sup {i_{n}}$ (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 extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined); and the assumption that one of the two hypotheses of the Statement holds.

[L1]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there, and the extended order is total (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 $R$ ). In particular $\inf {s_{n}} \leq s_{n}$ and $\sup {i_{n}} \geq i_{n}$ for every $n$ ; $q_{k} \leq s_{n}$ and $q_{k} \geq i_{n}$ for every $k \geq n$ ; a real $t$ exceeding $\inf {s_{n}}$ is not a lower bound of ${s_{n}}$ ; and a real $u$ below $\sup {i_{n}}$ is not an upper bound of ${i_{n}}$ .

[L2]

Both quantities exist for every sequence (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists 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 $\overline{R}$ ).

[L3]

Absolute value: $∣ x ∣ \geq 0$ ; $∣ x ∣ = 0$ exactly when $x = 0$ ; $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , so $∣ a_{k + 1} / a_{k} ∣ = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣$ (Basic properties of the absolute value).

[L4]

The principle of induction (The principle of mathematical induction).

[L5]

The geometric series $\sum_{j \geq 0} t^{j}$ converges when $∣ t ∣ < 1$ ; a series converges if and only if each of its tail series converges; and $\sum c x_{j}$ converges when $\sum x_{j}$ does (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Convergent series add and scale termwise).

[L6]

Direct comparison: if $0 \leq x_{j} \leq y_{j}$ from some index on and $\sum y_{j}$ converges then $\sum x_{j}$ converges (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 $\sum b_{k}$ ).

[L7]

If a series converges then its terms tend to $0$ (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals); and for every real $c > 0$ there is a natural $n \geq 1$ with the rational $1 / n < c$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L8]

Powers: $t^{0} = 1$ , $t^{j + 1} = t^{j} t$ , and $t^{j} > 0$ for $t > 0$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L9]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

Proof

technique · cases

1.1assume-case lt

Assume ${lim sup}_{k} q_{k} < 1$ .

1.2assume-case gt

Assume instead ${lim inf}_{k} q_{k} > 1$ .

1.3givenL1L2L3

Each $q_{k}$ is a nonnegative real, being a quotient of a nonnegative real by a positive one, so every $s_{n} \geq q_{n} \geq 0$ and hence ${lim sup}_{k} q_{k} \geq 0$ .

2.1step 1.1step 1.3L1choose

In the case ${lim sup}_{k} q_{k} < 1$ the value $Λ : = {lim sup}_{k} q_{k}$ therefore lies strictly between the reals $0$ and $1$ inclusive of $0$ , so it is a real number; put $t : = (Λ + 1) / 2$ , a real with $0 \leq Λ < t < 1$ .

2.2step 1.2L1choose

In the case ${lim inf}_{k} q_{k} > 1$ , the real $1$ is not an upper bound of ${i_{n}}$ , so there is $N \in N$ with $i_{N} > 1$ .

3.1step 2.1L1choose

In the case $Λ < 1$ : since $t > Λ = \inf {s_{n}}$ , the real $t$ is not a lower bound of ${s_{n}}$ , so there is $N$ with $s_{N} < t$ , and then $q_{k} \leq s_{N} < t$ for every $k \geq N$ .

3.2step 2.2L1

In the case ${lim inf}_{k} q_{k} > 1$ : $q_{k} \geq i_{N} > 1$ for every $k \geq N$ , since $i_{N}$ is a lower bound of ${q_{k} : k \geq N}$ .

3.3step 2.1L5

In the case $Λ < 1$ : the series $\sum_{j \geq 0} t^{j}$ converges since $∣ t ∣ = t < 1$ , hence so does $\sum_{j \geq 0} ∣ a_{N} ∣ t^{j}$ .

4.1step 3.1L3algebra

In the case $Λ < 1$ : for $k \geq N$ , $∣ a_{k + 1} ∣ / ∣ a_{k} ∣ < t$ with $∣ a_{k} ∣ > 0$ , hence $∣ a_{k + 1} ∣ < t ∣ a_{k} ∣$ .

4.2step 3.2L3algebra

In the case ${lim inf}_{k} q_{k} > 1$ : for $k \geq N$ , $∣ a_{k + 1} ∣ > ∣ a_{k} ∣$ , again multiplying by $∣ a_{k} ∣ > 0$ .

5.1step 4.1L4L8

In the case $Λ < 1$ : an induction on $j$ gives $∣ a_{N + j} ∣ \leq ∣ a_{N} ∣ t^{j}$ for every $j \in N$ . At $j = 0$ this is an equality, since $t^{0} = 1$ ; and if it holds at $j$ then $∣ a_{N + j + 1} ∣ < t ∣ a_{N + j} ∣ \leq t ∣ a_{N} ∣ t^{j} = ∣ a_{N} ∣ t^{j + 1}$ , using $t > 0$ .

5.2step 4.2L3L4

In the case ${lim inf}_{k} q_{k} > 1$ : an induction on $j$ gives $∣ a_{N + j} ∣ \geq ∣ a_{N} ∣ > 0$ for every $j \in N$ . At $j = 0$ it is an equality, and if it holds at $j$ then $∣ a_{N + j + 1} ∣ > ∣ a_{N + j} ∣ \geq ∣ a_{N} ∣$ .

6.1step 5.1step 3.3L3L5L6

In the case $Λ < 1$ : with $x_{j} : = ∣ a_{N + j} ∣$ and $y_{j} : = ∣ a_{N} ∣ t^{j}$ we have $0 \leq x_{j} \leq y_{j}$ for every $j$ , so $\sum_{j} ∣ a_{N + j} ∣$ converges; that is the $N$ -th tail series of $\sum ∣ a_{k} ∣$ , so $\sum ∣ a_{k} ∣$ converges.

6.2step 5.2L7choose

In the case ${lim inf}_{k} q_{k} > 1$ : $(a_{k})$ does not converge to $0$ . Choose a natural $n \geq 1$ with $1 / n < ∣ a_{N} ∣$ ; if $a_{k} \to 0$ there would be $K$ with $∣ a_{k} ∣ < 1 / n$ for all $k \geq K$ , contradicting $∣ a_{k} ∣ \geq ∣ a_{N} ∣ > 1 / n$ at any index $k$ that is at least both $K$ and $N$ .

7.1step 6.2L7

In the case ${lim inf}_{k} q_{k} > 1$ : by the term test $\sum a_{k}$ diverges, which is claim 2.

7.2step 6.1L9

In the case $Λ < 1$ : the series $\sum ∣ a_{k} ∣$ having been shown to converge, $\sum a_{k}$ converges as well; together with the convergence of $\sum ∣ a_{k} ∣$ that is claim 1.

8.1step 7.2step 7.1cases-exhaustive∎

The two assumed hypotheses are the cases of the disjunction in the Given, and they exhaust it; outside them both claims are vacuous, each hypothesis being false, so the theorem holds for every sequence with nonvanishing terms.

Remarks

The two halves are not dual, and the asymmetry is real. Convergence needs the ratios to be eventually below a fixed $t < 1$ , which $lim sup < 1$ supplies; divergence needs them eventually above $1$ , which is what $lim inf > 1$ supplies. A hypothesis on $lim sup$ alone can never force divergence, since a single large ratio occurring arbitrarily late says nothing about the size of the terms. That is exactly what FALSE: $lim sup ∣ a_{k + 1} / a_{k} ∣ \geq 1$ implies the series diverges records.
The geometric series is the only convergent series the proof knows. Claim 1 is a comparison against $∣ a_{N} ∣ t^{j}$ , and every later refinement on this page, Kummer's test included, exists because that comparison is too coarse when the ratios approach $1$ .

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Ratio test: lim sup⁡∣ak+1/ak∣<1limsup∣ak+1​/ak​∣<1 gives absolute convergence and hence convergence, and lim inf⁡∣ak+1/ak∣>1liminf∣ak+1​/ak​∣>1 gives divergence

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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