CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 $a_k \ne 0$ for every $k \in \mathbb{N}$ , and put

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

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

$\liminf_{k} q_k \;\le\; \liminf_{k} \rho_k \;\le\; \limsup_{k} \rho_k \;\le\; \limsup_{k} q_k ,$

and consequently:

if $\limsup_k q_k < 1$ , so that the ratio test gives convergence of $\sum |a_k|$ and hence of $\sum a_k$ , then $\limsup_k \rho_k < 1$ and the root test gives the same;
if $\liminf_k q_k > 1$ , so that the ratio test gives divergence of $\sum a_k$ , then $\limsup_k \rho_k > 1$ 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 $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , 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, $\limsup_k \rho_k = 1/2 < 1$ while $\limsup_k q_k = 2$ and $\liminf_k q_k = 1/8$ (FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ 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 $a_k \ne 0$ for every $k$ , the ratios $q_k = |a_{k+1}|/|a_k|$ and the roots $\rho_k = |a_{k+1}|^{1/(k+1)}$ (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).

[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 $\liminf_k q'_k \le \liminf_k r_k \le \limsup_k r_k \le \limsup_k q'_k$ in $\overline{\mathbb{R}}$ (For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ ).

[L2]

$\liminf_k x_k \le \limsup_k x_k$ for every real sequence ( $\liminf x_k \le \limsup x_k$ for every real sequence).

[L3]

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

[L4]

The root test: for a family from $1$ , $\limsup_k |a_{k+1}|^{1/(k+1)} < 1$ gives convergence of $\sum_{k \ge 1}|a_k|$ and hence of $\sum_{k \ge 1} a_k$ , and $> 1$ gives divergence of $\sum_{k \ge 1} a_k$ (Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing).

[L5]

The ratio test: $\limsup_k q_k < 1$ gives convergence of $\sum |a_k|$ and hence of $\sum a_k$ , and $\liminf_k q_k > 1$ gives divergence of $\sum a_k$ (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).

[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: $\limsup_k |a_{k+1}|^{1/(k+1)} = 1/2$ , $\limsup_k |a_{k+1}|/|a_k| = 2$ and $\liminf_k |a_{k+1}|/|a_k| = 1/8$ ; and $2^{-k} > 0$ , $t_k > 0$ , so every term is positive and in particular nonzero (FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ for every positive sequence, The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , 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 $a_k \ne 0$ 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, $\limsup_k \rho_k = 1/2 < 1$ , and neither $\limsup_k q_k = 2 < 1$ nor $\liminf_k q_k = 1/8 > 1$ 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 $b_{k+1}^{1/(k+1)} = |a_{k+1}|^{1/(k+1)} = \rho_k$ .

step 1.1

3.1

Therefore $\liminf_k q_k \le \liminf_k \rho_k \le \limsup_k \rho_k \le \limsup_k q_k$ , which is the displayed chain.

step 1.1step 2.1L1

4.1

Suppose $\limsup_k q_k < 1$ . By the chain, $\limsup_k \rho_k \le \limsup_k q_k < 1$ , so the root test applies to the family $(a_k)_{k \ge 1}$ and gives convergence of $\sum_{k \ge 1}|a_k|$ and of $\sum_{k \ge 1} a_k$ , 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 $\liminf_k q_k > 1$ . By the chain and [L2], $\limsup_k \rho_k \ge \liminf_k \rho_k \ge \liminf_k q_k > 1$ , so the root test gives divergence of $\sum_{k \ge 1} a_k$ , 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 $\sum_{k \ge 1} |a_k|$ 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 $\limsup$ , not about series. The whole content is the chain of For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 133 results over 34 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

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)