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

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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.

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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

1.1

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

assume-case lt

1.2

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

assume-case gt

1.3

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 ${lim sup}_{k} q_{k} \geq 0$ .

givenL1L2L3

2.1

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 $0 \leq Λ < t < 1$ .

step 1.1step 1.3L1choose

2.2

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 $i_{N} > 1$ .

step 1.2L1choose

3.1

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 $k \geq N$ .

step 2.1L1choose

3.2

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 ${q_{k} : k \geq N}$ .

step 2.2L1

3.3

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

step 2.1L5

4.1

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

step 3.1L3algebra

4.2

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

step 3.2L3algebra

5.1

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 $t > 0$ .

step 4.1L4L8

5.2

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 $∣ a_{N + j + 1} ∣ > ∣ a_{N + j} ∣ \geq ∣ a_{N} ∣$ .

step 4.2L3L4

6.1

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.

step 5.1step 3.3L3L5L6

6.2

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 $N$ .

step 5.2L7choose

7.1

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

step 6.2L7

7.2

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.

step 6.1L9

8.1

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.

step 7.2step 7.1cases-exhaustive∎

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 $1$ .

Depends on

Used by

Dependency tree · two levels

61 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

Ratio test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.34) (standard reference, not scraped)
Convergence tests (Wikipedia) (standard reference, not scraped)

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