TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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: $\limsup |a_{k+1}/a_k| < 1$ gives absolute convergence and hence convergence, and $\liminf |a_{k+1}/a_k| > 1$ gives divergence

Statement

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

$q_k \;:=\; \Big|\frac{a_{k+1}}{a_k}\Big| \;=\; \frac{|a_{k+1}|}{|a_k|} \qquad (k \in \mathbb{N}),$

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

if $\limsup_{k} q_k < 1$ then $\sum |a_k|$ converges, and hence $\sum a_k$ converges as well (If $\sum |a_k|$ converges then $\sum a_k$ converges);
if $\liminf_{k} q_k > 1$ then $\sum a_k$ diverges.

The hypothesis $a_k \ne 0$ 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 $\limsup_k q_k \ge 1 \ge \liminf_k q_k$ . 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 $a_k \ne 0$ for every $k$ ; the ratios $q_k = |a_{k+1}|/|a_k|$ ; the tail bounds $s_n = \sup\{q_k : k \ge n\}$ and $i_n = \inf\{q_k : k \ge n\}$ taken in $\overline{\mathbb{R}}$ , so that $\limsup_k q_k = \inf\{s_n\}$ and $\liminf_k q_k = \sup\{i_n\}$ (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 extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , 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 $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound there, and the extended order is total (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 $\mathbb{R}$ ). In particular $\inf\{s_n\} \le s_n$ and $\sup\{i_n\} \ge i_n$ for every $n$ ; $q_k \le s_n$ and $q_k \ge i_n$ for every $k \ge n$ ; 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 $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, 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 $\overline{\mathbb{R}}$ ).

[L3]

Absolute value: $|x| \ge 0$ ; $|x| = 0$ exactly when $x = 0$ ; $|xy| = |x||y|$ , 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 $\sum_{j \ge 0} t^{j}$ 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 $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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 $0 \le x_j \le y_j$ from some index on and $\sum y_j$ converges then $\sum x_j$ converges (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 $\sum b_k$ ).

[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 $n \ge 1$ with the rational $1/n < c$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[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 $\limsup_{k} q_k < 1$ .

assume-case lt

1.2

Assume instead $\liminf_{k} q_k > 1$ .

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 $s_n \ge q_n \ge 0$ and hence $\limsup_k q_k \ge 0$ .

givenL1L2L3

2.1

In the case $\limsup_k q_k < 1$ the value $\Lambda := \limsup_k q_k$ therefore lies strictly between the reals $0$ and $1$ inclusive of $0$ , so it is a real number; put $t := (\Lambda + 1)/2$ , a real with $0 \le \Lambda < t < 1$ .

step 1.1step 1.3L1choose

2.2

In the case $\liminf_k q_k > 1$ , the real $1$ is not an upper bound of $\{i_n\}$ , so there is $N \in \mathbb{N}$ with $i_N > 1$ .

step 1.2L1choose

3.1

In the case $\Lambda < 1$ : since $t > \Lambda = \inf\{s_n\}$ , the real $t$ is not a lower bound of $\{s_n\}$ , so there is $N$ with $s_N < t$ , and then $q_k \le s_N < t$ for every $k \ge N$ .

step 2.1L1choose

3.2

In the case $\liminf_k q_k > 1$ : $q_k \ge i_N > 1$ for every $k \ge N$ , since $i_N$ is a lower bound of $\{q_k : k \ge N\}$ .

step 2.2L1

3.3

In the case $\Lambda < 1$ : the series $\sum_{j \ge 0} t^{\,j}$ converges since $|t| = t < 1$ , hence so does $\sum_{j \ge 0} |a_N| t^{\,j}$ .

step 2.1L5

4.1

In the case $\Lambda < 1$ : for $k \ge N$ , $|a_{k+1}|/|a_k| < t$ with $|a_k| > 0$ , hence $|a_{k+1}| < t\,|a_k|$ .

step 3.1L3algebra

4.2

In the case $\liminf_k q_k > 1$ : for $k \ge N$ , $|a_{k+1}| > |a_k|$ , again multiplying by $|a_k| > 0$ .

step 3.2L3algebra

5.1

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

step 4.1L4L8

5.2

In the case $\liminf_k q_k > 1$ : an induction on $j$ gives $|a_{N+j}| \ge |a_N| > 0$ for every $j \in \mathbb{N}$ . At $j = 0$ it is an equality, and if it holds at $j$ then $|a_{N+j+1}| > |a_{N+j}| \ge |a_N|$ .

step 4.2L3L4

6.1

In the case $\Lambda < 1$ : with $x_j := |a_{N+j}|$ and $y_j := |a_N| t^{\,j}$ we have $0 \le x_j \le y_j$ 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 $\liminf_k q_k > 1$ : $(a_k)$ does not converge to $0$ . Choose a natural $n \ge 1$ with $1/n < |a_N|$ ; if $a_k \to 0$ there would be $K$ with $|a_k| < 1/n$ for all $k \ge K$ , contradicting $|a_k| \ge |a_N| > 1/n$ at any index $k$ that is at least both $K$ and $N$ .

step 5.2L7choose

7.1

In the case $\liminf_k q_k > 1$ : by the term test $\sum a_k$ diverges, which is claim 2.

step 6.2L7

7.2

In the case $\Lambda < 1$ : 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 $\limsup < 1$ supplies; divergence needs them eventually above $1$ , which is what $\liminf > 1$ supplies. A hypothesis on $\limsup$ 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: $\limsup |a_{k+1}/a_k| \ge 1$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 109 results over 27 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

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∣<1\limsup |a_{k+1}/a_k| < 1limsup∣ak+1​/ak​∣<1 gives absolute convergence and hence convergence, and lim inf⁡∣ak+1/ak∣>1\liminf |a_{k+1}/a_k| > 1liminf∣ak+1​/ak​∣>1 gives divergence