TheoremStatement: AI-adaptedProof: 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.

For $a_{k}, b_{k} > 0$ with $a_{k} / b_{k} \to L$ : if $L \in (0, \infty)$ the two series share their behaviour, while $L = 0$ and $L = \infty$ give one implication each

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals with $a_{k} > 0$ and $b_{k} > 0$ for every $k \in N$ , and put $q_{k} : = a_{k} / b_{k}$ . Then:

if $(q_{k})$ converges with $\lim_{k} q_{k} = L$ for a real $L > 0$ (Limits and Cauchy sequences of reals), then $\sum a_{k}$ converges if and only if $\sum b_{k}$ converges;
if $(q_{k})$ converges with $\lim_{k} q_{k} = 0$ , then convergence of $\sum b_{k}$ implies convergence of $\sum a_{k}$ ; equivalently, divergence of $\sum a_{k}$ implies divergence of $\sum b_{k}$ ;
if $(q_{k})$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ), then convergence of $\sum a_{k}$ implies convergence of $\sum b_{k}$ ; equivalently, divergence of $\sum b_{k}$ implies divergence of $\sum a_{k}$ .

In each clause the convergence of $(q_{k})$ , or its divergence to $+ \infty$ , is part of the hypothesis, so the symbol $\lim_{k} q_{k}$ denotes wherever it is written (A sequence has at most one limit).

Neither implication in claim 2 can be reversed, and by symmetry neither can the one in claim 3; the companion page exhibits a pair with $\lim_{k} q_{k} = 0$ , $\sum a_{k}$ convergent and $\sum b_{k}$ divergent.

For families from a general starting index $m$ the statement is the same, applied to the shifted sequences $j \mapsto a_{j + m}$ and $j \mapsto b_{j + m}$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

On the third regime. " $L = \infty$ " is written here as divergence of $(q_{k})$ to $+ \infty$ in the sense of Divergence to $+ \infty$ and to $- \infty$ , and never as a limit equation with an infinite right-hand side. A sequence diverging to $+ \infty$ has no limit in $R$ , and this library does not write $\lim q_{k} = + \infty$ .

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(b_{k})$ of reals with $a_{k} > 0$ and $b_{k} > 0$ for every $k$ , the quotients $q_{k} = a_{k} / b_{k}$ , and the assumption that one of the three regimes of the Statement holds: $(q_{k})$ converges with $\lim_{k} q_{k} = L$ for some real $L > 0$ ; or $(q_{k})$ converges with $\lim_{k} q_{k} = 0$ ; or $(q_{k})$ diverges to $+ \infty$ (A sequence has at most one limit).

[L1]

Convergence to $x$ means: for every rational $ε > 0$ there is $K$ with $∣ x_{k} - x ∣ < ε$ for all $k \geq K$ ; and the same holds for every real $ε > 0$ , since every real $ε > 0$ exceeds some rational $1 / n$ with $n \geq 1$ natural (Limits and Cauchy sequences of reals, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L2]

$q_{k} \to + \infty$ means: for every real $M$ there is $K$ with $q_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[L3]

Direct comparison: if $0 \leq x_{k} \leq y_{k}$ for all $k$ from some index on, then convergence of $\sum y_{k}$ gives convergence of $\sum x_{k}$ (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 $\sum b_{k}$ ).

[L4]

For $c \neq 0$ : $\sum c x_{k}$ converges if and only if $\sum x_{k}$ converges (Convergent series add and scale termwise).

[L5]

Since $q_{k} = a_{k} / b_{k}$ and $b_{k} > 0$ , the field laws give $a_{k} = q_{k} b_{k}$ . Multiplication by a positive scalar preserves strict inequalities; the non-strict form follows by adjoining the equality case (Field, Sign rules for products and monotonicity of multiplication).

Proof

technique · cases

1.1

Assume $(q_{k})$ converges with $\lim_{k} q_{k} = L$ for a real $L > 0$ .

assume-case pos

1.2

Assume instead $(q_{k})$ converges with $\lim_{k} q_{k} = 0$ .

assume-case zero

1.3

Assume instead $(q_{k})$ diverges to $+ \infty$ .

assume-case inf

2.1

In the case $\lim_{k} q_{k} = L > 0$ , apply [L1] with the real tolerance $L / 2 > 0$ : there is $K$ with $∣ q_{k} - L ∣ < L / 2$ , hence $L / 2 < q_{k} < 3 L / 2$ , for all $k \geq K$ .

step 1.1L1choose

2.2

In the case $\lim_{k} q_{k} = 0$ , apply [L1] with the rational tolerance $1$ : there is $K$ with $∣ q_{k} ∣ < 1$ , hence $q_{k} < 1$ , for all $k \geq K$ .

step 1.2L1choose

2.3

In the case $q_{k} \to + \infty$ , apply [L2] with $M = 1$ : there is $K$ with $q_{k} > 1$ for all $k \geq K$ .

step 1.3L2choose

3.1

In the case $\lim_{k} q_{k} = L > 0$ , multiplying by $b_{k} > 0$ turns step 2.1 into $(L / 2) b_{k} < a_{k} < (3 L / 2) b_{k}$ for all $k \geq K$ , and all three quantities are positive.

step 2.1L5algebra

3.2

In the case $\lim_{k} q_{k} = 0$ , multiplying by $b_{k} > 0$ turns step 2.2 into $0 < a_{k} < b_{k}$ for all $k \geq K$ .

step 2.2L5algebra

3.3

In the case $q_{k} \to + \infty$ , multiplying by $b_{k} > 0$ turns step 2.3 into $0 < b_{k} < a_{k}$ for all $k \geq K$ .

step 2.3L5algebra

4.1

In the case $L > 0$ : if $\sum b_{k}$ converges then so does $\sum (3 L / 2) b_{k}$ , and $0 \leq a_{k} \leq (3 L / 2) b_{k}$ for $k \geq K$ , so $\sum a_{k}$ converges.

step 3.1L3L4

4.2

In the case $L > 0$ : if $\sum a_{k}$ converges then, since $0 \leq (L / 2) b_{k} \leq a_{k}$ for $k \geq K$ , the series $\sum (L / 2) b_{k}$ converges, and $L / 2 \neq 0$ gives convergence of $\sum b_{k}$ .

step 3.1L3L4

4.3

In the case $\lim_{k} q_{k} = 0$ : $0 \leq a_{k} \leq b_{k}$ for $k \geq K$ , so convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and the contrapositive is the divergence form.

step 3.2L3

4.4

In the case $q_{k} \to + \infty$ : $0 \leq b_{k} \leq a_{k}$ for $k \geq K$ , so convergence of $\sum a_{k}$ gives convergence of $\sum b_{k}$ , and the contrapositive is the divergence form.

step 3.3L3

5.1

The two implications in the case $L > 0$ are the two directions of claim 1, and the remaining two cases give claims 2 and 3. The three assumed regimes are the cases of the disjunction in the Given, and they exhaust it, so every instance of the theorem is covered: outside those three regimes each of the three implications is vacuous, its hypothesis being false.

step 4.1step 4.2step 4.3step 4.4cases-exhaustive∎

Remarks

Why the three regimes are treated as one proof. The Statement is a conjunction of three implications, each with its own hypothesis on $(q_{k})$ . Fixing the two sequences and arguing by cases on which regime holds proves all three at once, and costs nothing: if none of the regimes holds, every one of the three implications is vacuously true.
Positivity of $(b_{k})$ is needed twice. It is what makes $q_{k}$ defined at all, and it is what lets an inequality between the $q_{k}$ be multiplied through to an inequality between the $a_{k}$ and the $b_{k}$ without reversing. Positivity of $(a_{k})$ is what supplies the lower bound $0$ that the direct comparison test requires.
The limit is only used through an eventual two-sided estimate. No step needs the exact value of $L$ , only that $q_{k}$ is eventually trapped strictly between two positive multiples of it. That is why the test still works when the quotients merely stay between two positive constants, and why the hypothesis $\lim_{k} q_{k} = L$ is stronger than what the proof consumes.

Depends on

Used by

With aₖ/bₖ → 0, convergence of ∑ aₖ does not give convergence of ∑ bₖ Counterexample

Dependency tree · two levels

33 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

Limit comparison test (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §7.3 (standard reference, not scraped)
APEX Calculus, Section 9.4: Comparison Tests (standard reference, not scraped)
MIT 18.100B Real Analysis lecture notes (standard reference, not scraped)

For ak,bk>0ak​,bk​>0 with ak/bk→Lak​/bk​→L: if L∈(0,∞)L∈(0,∞) the two series share their behaviour, while L=0L=0 and L=∞L=∞ give one implication each