TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$

Statement

Let $(a_k)_{k \in \mathbb{N}}$ be a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with $a_k > 0$ for every $k$ . Put

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

with roots as in Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ and Rational powers $a^r$ of a positive base. Then, in $\overline{\mathbb{R}}$ (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),

$\liminf_{k} q_k \;\le\; \liminf_{k} r_k \;\le\; \limsup_{k} r_k \;\le\; \limsup_{k} q_k .$

The root sequence must start at index $1$ , and $(r_k)$ is the shift that makes it a sequence on $\mathbb{N}$ . The classical statement writes $a_n^{1/n}$ , which is meaningful only for $n \ge 1$ , since $1/0$ is not a rational number; sequences here are functions on $\mathbb{N}$ and $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so the root family is written $r_k = a_{k+1}^{1/(k+1)}$ , which is $a_n^{1/n}$ reindexed by $n = k+1$ . The ratio family $q_k$ needs no shift, and the four quantities in the display are those of the two sequences $(q_k)$ and $(r_k)$ exactly as written here.

This is why the root test dominates the ratio test. If the ratios converge, the outer two quantities coincide and the chain forces the roots to converge to the same value; but the roots can converge when the ratios do not, and then the chain is strict at both ends. Both phenomena are exhibited by named examples on the companion page.

Facts & Assumptions

Given: A sequence $(a_k)$ of reals with $a_k > 0$ for every $k$ ; the ratio sequence $q_k = a_{k+1}/a_k$ ; the root sequence $r_k = a_{k+1}^{1/(k+1)}$ ; and $\iota(n) = n \cdot 1_{\mathbb{R}}$ for the canonical naturals.

[L1]

Tail ranges, extended tail bounds and the two quantities exist for every sequence, with $s_n$ the least upper bound of the $n$ -th tail range and $i_n$ its greatest lower bound, $\limsup = \inf\{s_n\}$ and $\liminf = \sup\{i_n\}$ (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 tail suprema of any real sequence are nonincreasing in $\overline{\mathbb{R}}$ , so the limit superior exists for every sequence, 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 $\mathbb{R}$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

The order on $\overline{\mathbb{R}}$ is total and transitive, $+\infty$ is greatest and $-\infty$ least, it restricts on $\mathbb{R}$ to the order of $\mathbb{R}$ , and an element between two reals is real (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined, Partial order and partially ordered set).

[L3]

Epsilon characterisation, for a real $L$ : $L = \limsup_k z_k$ gives $z_k < L + \varepsilon$ eventually for every real $\varepsilon > 0$ ; $L = \liminf_k z_k$ gives $z_k > L - \varepsilon$ eventually for every real $\varepsilon > 0$ (For finite $L$ : $L = \limsup x_k$ iff for every $\varepsilon > 0$ one has $x_k < L + \varepsilon$ eventually and $x_k > L - \varepsilon$ frequently).

[L4]

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

[L5]

Comparison: $z_k \le w_k$ eventually implies $\limsup_k z_k \le \limsup_k w_k$ and $\liminf_k z_k \le \liminf_k w_k$ (If $x_k \le y_k$ eventually then $\limsup x_k \le \limsup y_k$ and $\liminf x_k \le \liminf y_k$ ).

[L6]

A sequence converging to a real $c$ has $\limsup = \liminf = c$ ; and $\liminf_k z_k = +\infty$ implies $z_k \to +\infty$ , hence $z_k > M$ eventually for every real $M$ (A real sequence converges to $L \in \mathbb{R}$ iff $\liminf x_k = \limsup x_k = L$ , and diverges to $\pm\infty$ iff both equal $\pm\infty$ , Divergence to $+\infty$ and to $-\infty$ ).

[L7]

For every real $C > 0$ the sequence $C^{1/(k+1)}$ converges to $1$ (For every $a > 0$ , $a^{1/n} \to 1$ ).

[L8]

Algebra of limits: a scalar multiple of a convergent sequence converges to the scalar multiple of the limit (Algebra of limits: sums, scalar multiples, products and quotients).

[L9]

Roots and powers of positive reals: $x^{1/n}$ exists, is unique and is $> 0$ for $x > 0$ and $n \ge 1$ ; $(xy)^{1/n} = x^{1/n} y^{1/n}$ ; the integer power $x^n$ is the rational power at exponent $n$ , so $(x^n)^{1/n} = x^{n \cdot (1/n)} = x$ ; $x^{-m} = 1/x^m$ and $x^{m} x^{m'} = x^{m+m'}$ for integer exponents and $x \ne 0$ ; $x^n > 0$ for $x > 0$ ; and $0 \le x \le y$ implies $x^{1/n} \le y^{1/n}$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ , Rational powers $a^r$ of a positive base, Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Integer powers $a^m$ , Laws of integer exponents, Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ ).

[L10]

Induction principle (The principle of mathematical induction).

[L11]

Archimedean facts: for every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ ; and $0 < x < y$ gives $0 < 1/y < 1/x$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L12]

Order arithmetic: Order is preserved by adding a constant and by adding inequalities and claim 4 of Sign rules for products and monotonicity of multiplication state the strict forms, that inequalities may be translated and added and that multiplication by a positive element preserves $<$ ; adjoining the case of equality, where both sides move or scale alike, gives the nonstrict forms used below. Products of nonnegative inequalities multiply in the nonstrict form stated by Multiplying inequalities of positives, and the order on $\mathbb{R}$ is total.

[L13]

Strictly between any two reals lies a rational (The rationals embed densely in the reals).

[L14]

The order on $\mathbb{N}$ is total and transitive (Order on the natural numbers, $\le$ is a linear order on $\mathbb{N}$ , Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Every $q_k$ is positive, being a quotient of positive reals, and every $r_k$ is positive, being a root of the positive real $a_{k+1}$ . Hence $0$ is a lower bound of every tail range of $(q_k)$ and of $(r_k)$ , so every tail infimum is $\ge 0$ and therefore $\liminf_k q_k \ge 0$ and $\liminf_k r_k \ge 0$ ; with [L4] this also gives $\limsup_k q_k \ge 0$ .

givenL1L2L4L9L11

1.2

Let $c > 0$ be real, let $N \in \mathbb{N}$ and put $C := a_N c^{-N}$ , a positive real. If $a_{k+1} \le c\,a_k$ for every $k \ge N$ then $a_n \le C c^n$ for every $n \ge N$ ; if $a_{k+1} \ge c\,a_k$ for every $k \ge N$ then $a_n \ge C c^n$ for every $n \ge N$ . Both are inductions on $j$ for $n = N + j$ : at $j = 0$ one has $C c^N = a_N c^{-N} c^N = a_N c^0 = a_N$ , and the inductive step multiplies the bound at $n$ by the positive $c$ and uses the hypothesis at $k = n$ .

givenL9L10L12

1.3

Let $C > 0$ and $c > 0$ be real and $n \ge 1$ a natural. Then $(C c^n)^{1/n} = C^{1/n} (c^n)^{1/n} = C^{1/n} c$ . Consequently $0 < a_n \le C c^n$ gives $a_n^{1/n} \le C^{1/n} c$ , and $a_n \ge C c^n > 0$ gives $a_n^{1/n} \ge C^{1/n} c$ , since $x \mapsto x^{1/n}$ is nondecreasing on the nonnegative reals.

givenL9

1.4

For real $C > 0$ and $c > 0$ the sequence $u_k := C^{1/(k+1)} c$ converges to $c$ , by [L7] and the scalar rule; hence $\limsup_k u_k = \liminf_k u_k = c$ .

givenL6L7L8

1.5

If $\limsup_k q_k = +\infty$ then $\limsup_k r_k \le \limsup_k q_k$ , since $+\infty$ is the greatest element of $\overline{\mathbb{R}}$ .

givenL2

2.1

Suppose $\beta := \limsup_k q_k$ is real, and let $\varepsilon > 0$ be an arbitrary real. Put $c := \beta + \varepsilon$ , which is positive since $\beta \ge 0$ . By [L3] there is $N$ with $q_k < c$ for all $k \ge N$ , that is $a_{k+1} < c\,a_k$ after multiplying by $a_k > 0$ ; so $a_{k+1} \le c\,a_k$ for $k \ge N$ , and step 1.2 gives $a_n \le C c^n$ for all $n \ge N$ with $C := a_N c^{-N} > 0$ . For $k \ge N$ the index $n := k+1$ satisfies $n \ge N$ and $n \ge 1$ , so step 1.3 gives $r_k \le C^{1/(k+1)} c = u_k$ . By step 1.4 and [L5], $\limsup_k r_k \le \limsup_k u_k = c = \beta + \varepsilon$ .

step 1.1step 1.2step 1.3step 1.4L3L5L12L14

2.2

If $\alpha := \liminf_k q_k = 0$ then $\liminf_k r_k \ge 0 = \alpha$ by step 1.1.

step 1.1

2.3

Suppose $\alpha := \liminf_k q_k > 0$ and let $c$ be a real with $0 < c < \alpha$ . Then $q_k > c$ eventually: if $\alpha$ is real this is [L3] applied with $\varepsilon := \alpha - c > 0$ , and if $\alpha = +\infty$ then $q_k \to +\infty$ by [L6], so $q_k > c$ eventually. Fix $N$ with $q_k > c$ for all $k \ge N$ ; then $a_{k+1} \ge c\,a_k$ for $k \ge N$ , so step 1.2 gives $a_n \ge C c^n$ for all $n \ge N$ with $C := a_N c^{-N} > 0$ , and step 1.3 gives $r_k \ge C^{1/(k+1)} c = u_k$ for every $k \ge N$ . By step 1.4 and [L5], $\liminf_k r_k \ge \liminf_k u_k = c$ .

step 1.1step 1.2step 1.3step 1.4L3L5L6L12L14

3.1

Hence $\limsup_k r_k \le \limsup_k q_k$ . By step 1.1 the element $\beta = \limsup_k q_k$ is $\ge 0$ , so it is either $+\infty$ , which is step 1.5, or real. In the real case step 2.1 with $\varepsilon = 1$ gives $\limsup_k r_k \le \beta + 1$ , a real, so $\limsup_k r_k \ne +\infty$ ; if $\limsup_k r_k = -\infty$ it is $\le \beta$ ; and otherwise it is a real $S$ , and $S > \beta$ would give, on choosing a natural $m \ge 1$ with $1/m < S - \beta$ and applying step 2.1 with $\varepsilon = 1/m$ , the impossibility $S \le \beta + 1/m < S$ . By totality $\limsup_k r_k \le \beta$ .

step 2.1step 1.5step 1.1L2L11L12

3.2

Hence $\liminf_k q_k \le \liminf_k r_k$ . By step 1.1 the element $\alpha = \liminf_k q_k$ is $\ge 0$ , so it is $0$ , or a positive real, or $+\infty$ . The first case is step 2.2. If $\alpha$ is a positive real and $\liminf_k r_k < \alpha$ , then $\liminf_k r_k$ lies between the reals $0$ and $\alpha$ by step 1.1 and is therefore real, so [L13] supplies a real $c$ with $\liminf_k r_k < c < \alpha$ , necessarily $c > 0$ ; step 2.3 then gives $\liminf_k r_k \ge c$ , contradicting $c > \liminf_k r_k$ , so $\liminf_k r_k \ge \alpha$ by totality. If $\alpha = +\infty$ , step 2.3 gives $\liminf_k r_k \ge c$ for every real $c$ with $c > 0$ , so $\liminf_k r_k$ is not $-\infty$ , and it is not a real $t$ either, since $t \ge 0$ by step 1.1 and then $c := t+1 > 0$ would give $t \ge t+1$ ; hence $\liminf_k r_k = +\infty = \alpha$ .

step 2.2step 2.3step 1.1L2L12L13

4.1

Combining the three links, $\liminf_k q_k \le \liminf_k r_k$ by step 3.2, $\liminf_k r_k \le \limsup_k r_k$ by [L4], and $\limsup_k r_k \le \limsup_k q_k$ by step 3.1.

step 3.1step 3.2L4∎

Remarks

The mechanism is that a ratio bound integrates to a geometric bound. If the ratios are eventually below $c$ then the terms are eventually below a constant times $c^n$ , and taking $n$ -th roots turns the constant into $C^{1/n}$ , which tends to $1$ by For every $a > 0$ , $a^{1/n} \to 1$ . That single lemma is what makes the constant disappear, and it is the only analytic input; everything else is the comparison lemma If $x_k \le y_k$ eventually then $\limsup x_k \le \limsup y_k$ and $\liminf x_k \le \liminf y_k$ and order bookkeeping.
All four quantities can be different, and the two outer inequalities can both be strict. A positive sequence making all three inequalities of the ratio-to-root chain strict ↗ gives a positive sequence with chain $0 < 1/3 < 1/2 < +\infty$ , so no two of the four coincide, and $a_k = 2^{-k + (-1)^k}$ has $\liminf a_{k+1}/a_k = 1/8$ , $\limsup a_{k+1}/a_k = 2$ and $\lim a_k^{1/k} = 1/2$ ↗ is the standard witness in which the roots converge while the ratios oscillate across $1$ .
The chain also explains the practical rule. When $\limsup_k q_k < 1$ the chain forces $\limsup_k r_k < 1$ , so any conclusion drawn from the roots is available from the ratios; but $\limsup_k r_k < 1$ can hold with $\limsup_k q_k > 1$ , and then only the root side is usable. This is the sense in which the root criterion is the stronger of the two.
The hypothesis $a_k > 0$ is needed at every index, not merely eventually. The ratios must be defined, which needs $a_k \ne 0$ , and the roots must be defined, which needs $a_{k+1} \ge 0$ ; positivity also lets the ratio inequalities be cleared of denominators. A sequence positive only from some index on can be handled by passing to that tail, which changes none of the four quantities.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 127 results over 40 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)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.37) (standard reference, not scraped)
University of Maryland Math 410, Notes on the Real Numbers (standard reference, not scraped)

For ak>0a_k > 0ak​>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_kliminfak+1​/ak​≤liminfak1/k​≤limsupak1/k​≤limsupak+1​/ak​