CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Whenever the ratio test decides, the root test decides the same way, and the converse fails

Statement

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

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

the ratio and root families of 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 and Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing. Then, in $\overline{R}$ ,

${lim inf}_{k} q_{k} \leq {lim inf}_{k} ρ_{k} \leq {lim sup}_{k} ρ_{k} \leq {lim sup}_{k} q_{k},$

and consequently:

if ${lim sup}_{k} q_{k} < 1$ , so that the ratio test gives convergence of $\sum ∣ a_{k} ∣$ and hence of $\sum a_{k}$ , then ${lim sup}_{k} ρ_{k} < 1$ and the root test gives the same;
if ${lim inf}_{k} q_{k} > 1$ , so that the ratio test gives divergence of $\sum a_{k}$ , then ${lim sup}_{k} ρ_{k} > 1$ and the root test gives it too.

The converse fails. Let $(s_{k})$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , let $t_{k} : = 2$ when $s_{k} = 1$ and $t_{k} : = 1 / 2$ when $s_{k} = - 1$ , and put $a_{k} : = 2^{- k} t_{k}$ , the sequence usually written $a_{k} = 2^{- k + (- 1)^{k}}$ . For it, ${lim sup}_{k} ρ_{k} = 1 / 2 < 1$ while ${lim sup}_{k} q_{k} = 2$ and ${lim inf}_{k} q_{k} = 1 / 8$ (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence), so the root test gives convergence of $\sum ∣ a_{k} ∣$ and neither half of the ratio test applies. So the root test decides strictly more series than the ratio test.

Facts & Assumptions

Given: A sequence $(a_{k})$ of reals with $a_{k} \neq 0$ for every $k$ , the ratios $q_{k} = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣$ and the roots $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ (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).

[L1]

For a sequence $(b_{k})$ of reals with $b_{k} > 0$ for every $k$ , writing $q_{k}^{'} = b_{k + 1} / b_{k}$ and $r_{k} = b_{k + 1}^{1 / (k + 1)}$ , one has ${lim inf}_{k} q_{k}^{'} \leq {lim inf}_{k} r_{k} \leq {lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k}^{'}$ in $\overline{R}$ (For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ ).

[L2]

${lim inf}_{k} x_{k} \leq {lim sup}_{k} x_{k}$ for every real sequence ( $lim inf x_{k} \leq lim sup x_{k}$ for every real sequence).

[L3]

Absolute value: $∣ x ∣ \geq 0$ , and $∣ x ∣ = 0$ exactly when $x = 0$ (Basic properties of the absolute value).

[L4]

The root test: for a family from $1$ , ${lim sup}_{k} ∣ a_{k + 1} ∣^{1 / (k + 1)} < 1$ gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ and hence of $\sum_{k \geq 1} a_{k}$ , and $> 1$ gives divergence of $\sum_{k \geq 1} a_{k}$ (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L5]

The ratio test: ${lim sup}_{k} q_{k} < 1$ gives convergence of $\sum ∣ a_{k} ∣$ and hence of $\sum a_{k}$ , and ${lim inf}_{k} q_{k} > 1$ gives divergence of $\sum a_{k}$ (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).

[L6]

A series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

For the sequence $a_{k} = 2^{- k} t_{k}$ built from the alternating sequence as in the Statement: ${lim sup}_{k} ∣ a_{k + 1} ∣^{1 / (k + 1)} = 1 / 2$ , ${lim sup}_{k} ∣ a_{k + 1} ∣ / ∣ a_{k} ∣ = 2$ and ${lim inf}_{k} ∣ a_{k + 1} ∣ / ∣ a_{k} ∣ = 1 / 8$ ; and $2^{- k} > 0$ , $t_{k} > 0$ , so every term is positive and in particular nonzero (FALSE: $lim sup a_{k}^{1 / k} = lim sup a_{k + 1} / a_{k}$ for every positive sequence, The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ , Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

Proof

technique · direct

1.1

Put $b_{k} : = ∣ a_{k} ∣$ . Since $a_{k} \neq 0$ we have $b_{k} > 0$ for every $k$ , so [L1] applies to $(b_{k})$ .

givenL3L1

1.2

For the sequence $a_{k} = 2^{- k} t_{k}$ of the Statement every term is nonzero, ${lim sup}_{k} ρ_{k} = 1 / 2 < 1$ , and neither ${lim sup}_{k} q_{k} = 2 < 1$ nor ${lim inf}_{k} q_{k} = 1 / 8 > 1$ holds.

2.1

For this $(b_{k})$ the ratio family is $b_{k + 1} / b_{k} = ∣ a_{k + 1} ∣ / ∣ a_{k} ∣ = q_{k}$ and the root family is $b_{k + 1}^{1 / (k + 1)} = ∣ a_{k + 1} ∣^{1 / (k + 1)} = ρ_{k}$ .

step 1.1

3.1

Therefore ${lim inf}_{k} q_{k} \leq {lim inf}_{k} ρ_{k} \leq {lim sup}_{k} ρ_{k} \leq {lim sup}_{k} q_{k}$ , which is the displayed chain.

step 1.1step 2.1L1

4.1

Suppose ${lim sup}_{k} q_{k} < 1$ . By the chain, ${lim sup}_{k} ρ_{k} \leq {lim sup}_{k} q_{k} < 1$ , so the root test applies to the family $(a_{k})_{k \geq 1}$ and gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ and of $\sum_{k \geq 1} a_{k}$ , hence of $\sum ∣ a_{k} ∣$ and of $\sum a_{k}$ ; the ratio test gives the same conclusions. That is claim 1.

step 3.1L4L5L6

4.2

Suppose ${lim inf}_{k} q_{k} > 1$ . By the chain and [L2], ${lim sup}_{k} ρ_{k} \geq {lim inf}_{k} ρ_{k} \geq {lim inf}_{k} q_{k} > 1$ , so the root test gives divergence of $\sum_{k \geq 1} a_{k}$ , hence of $\sum a_{k}$ ; the ratio test gives the same conclusion. That is claim 2.

step 3.1L2L4L5L6

5.1

So for that sequence the root test gives convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ while neither half of the ratio test applies, and the converse of claims 1 and 2 fails.

step 1.2L4L5∎

Remarks

The dominance is a statement about $lim sup$ , not about series. The whole content is the chain of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ , proved on the previous page precisely because it is about limits superior and nothing else. Claims 1 and 2 are the translation of that chain through the two tests, and they carry no further mathematics.
Strictly more, not merely at least as much. The witness in the Statement settles that: its roots converge to $1 / 2$ while its ratios oscillate between $1 / 8$ and $2$ , so the ratio test is silent about a series the root test decides. The reason is structural rather than accidental. Taking an $n$ -th root divides the exponent by $n$ and so damps a bounded oscillation, while forming a ratio differences the exponent and preserves it.
The ratio test survives because it is easier to compute. Nothing here says the ratio test should be abandoned; the ratios of a series given by an explicit formula are usually elementary, and the roots usually are not.

Depends on

Used by

Dependency tree · two levels

76 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

Root test (Wikipedia) (standard reference, not scraped)
Ratio test (Wikipedia) (standard reference, not scraped)
CSUDH notes on the ratio and root tests (standard reference, not scraped)