TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$

Statement

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

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

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

${lim inf}_{k} q_{k} \leq {lim inf}_{k} r_{k} \leq {lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k} .$

The root sequence must start at index $1$ , and $(r_{k})$ is the shift that makes it a sequence on $N$ . The classical statement writes $a_{n}^{1 / n}$ , which is meaningful only for $n \geq 1$ , since $1 / 0$ is not a rational number; sequences here are functions on $N$ and $N$ 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 $ι (n) = n \cdot 1_{R}$ 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, $lim sup = \inf {s_{n}}$ and $lim inf = \sup {i_{n}}$ (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 tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ , Upper bound, least upper bound, and strict upper bound, Partial order and partially ordered set).

[L2]

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

[L3]

Epsilon characterisation, for a real $L$ : $L = {lim sup}_{k} z_{k}$ gives $z_{k} < L + ε$ eventually for every real $ε > 0$ ; $L = {lim inf}_{k} z_{k}$ gives $z_{k} > L - ε$ eventually for every real $ε > 0$ (For finite $L$ : $L = lim sup x_{k}$ iff for every $ε > 0$ one has $x_{k} < L + ε$ eventually and $x_{k} > L - ε$ frequently).

[L4]

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

[L5]

Comparison: $z_{k} \leq w_{k}$ eventually implies ${lim sup}_{k} z_{k} \leq {lim sup}_{k} w_{k}$ and ${lim inf}_{k} z_{k} \leq {lim inf}_{k} w_{k}$ (If $x_{k} \leq y_{k}$ eventually then $lim sup x_{k} \leq lim sup y_{k}$ and $lim inf x_{k} \leq lim inf y_{k}$ ).

[L6]

A sequence converging to a real $c$ has $lim sup = lim inf = c$ ; and ${lim inf}_{k} z_{k} = + \infty$ implies $z_{k} \to + \infty$ , hence $z_{k} > M$ eventually for every real $M$ (A real sequence converges to $L \in R$ iff $lim inf x_{k} = lim sup x_{k} = L$ , 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 $n \geq 1$ ; $(x y)^{1 / n} = x^{1 / n} y^{1 / n}$ ; 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 $x \neq 0$ ; $x^{n} > 0$ for $x > 0$ ; and $0 \leq x \leq y$ implies $x^{1 / n} \leq y^{1 / n}$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ 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 $η > 0$ there is a natural $m \geq 1$ with $1 / m < η$ ; and $0 < x < y$ gives $0 < 1 / y < 1 / x$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , 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 $R$ is total.

[L13]

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

[L14]

The order on $N$ is total and transitive (Order on the natural numbers, $\leq$ is a linear order on $N$ , 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 $\geq 0$ and therefore ${lim inf}_{k} q_{k} \geq 0$ and ${lim inf}_{k} r_{k} \geq 0$ ; with [L4] this also gives ${lim sup}_{k} q_{k} \geq 0$ .

givenL1L2L4L9L11

1.2

Let $c > 0$ be real, let $N \in N$ and put $C : = a_{N} c^{- N}$ , a positive real. If $a_{k + 1} \leq c a_{k}$ for every $k \geq N$ then $a_{n} \leq C c^{n}$ for every $n \geq N$ ; if $a_{k + 1} \geq c a_{k}$ for every $k \geq N$ then $a_{n} \geq C c^{n}$ for every $n \geq N$ . 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 $n \geq 1$ a natural. Then $(C c^{n})^{1 / n} = C^{1 / n} (c^{n})^{1 / n} = C^{1 / n} c$ . Consequently $0 < a_{n} \leq C c^{n}$ gives $a_{n}^{1 / n} \leq C^{1 / n} c$ , and $a_{n} \geq C c^{n} > 0$ gives $a_{n}^{1 / n} \geq C^{1 / n} c$ , 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 ${lim sup}_{k} u_{k} = {lim inf}_{k} u_{k} = c$ .

givenL6L7L8

1.5

If ${lim sup}_{k} q_{k} = + \infty$ then ${lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k}$ , since $+ \infty$ is the greatest element of $\overline{R}$ .

givenL2

2.1

Suppose $β : = {lim sup}_{k} q_{k}$ is real, and let $ε > 0$ be an arbitrary real. Put $c : = β + ε$ , which is positive since $β \geq 0$ . By [L3] there is $N$ with $q_{k} < c$ for all $k \geq N$ , that is $a_{k + 1} < c a_{k}$ after multiplying by $a_{k} > 0$ ; so $a_{k + 1} \leq c a_{k}$ for $k \geq N$ , and step 1.2 gives $a_{n} \leq C c^{n}$ for all $n \geq N$ with $C : = a_{N} c^{- N} > 0$ . For $k \geq N$ the index $n : = k + 1$ satisfies $n \geq N$ and $n \geq 1$ , so step 1.3 gives $r_{k} \leq C^{1 / (k + 1)} c = u_{k}$ . By step 1.4 and [L5], ${lim sup}_{k} r_{k} \leq {lim sup}_{k} u_{k} = c = β + ε$ .

step 1.1step 1.2step 1.3step 1.4L3L5L12L14

2.2

If $α : = {lim inf}_{k} q_{k} = 0$ then ${lim inf}_{k} r_{k} \geq 0 = α$ by step 1.1.

step 1.1

2.3

Suppose $α : = {lim inf}_{k} q_{k} > 0$ and let $c$ be a real with $0 < c < α$ . Then $q_{k} > c$ eventually: if $α$ is real this is [L3] applied with $ε : = α - c > 0$ , and if $α = + \infty$ then $q_{k} \to + \infty$ by [L6], so $q_{k} > c$ eventually. Fix $N$ with $q_{k} > c$ for all $k \geq N$ ; then $a_{k + 1} \geq c a_{k}$ for $k \geq N$ , so step 1.2 gives $a_{n} \geq C c^{n}$ for all $n \geq N$ with $C : = a_{N} c^{- N} > 0$ , and step 1.3 gives $r_{k} \geq C^{1 / (k + 1)} c = u_{k}$ for every $k \geq N$ . By step 1.4 and [L5], ${lim inf}_{k} r_{k} \geq {lim inf}_{k} u_{k} = c$ .

step 1.1step 1.2step 1.3step 1.4L3L5L6L12L14

3.1

Hence ${lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k}$ . By step 1.1 the element $β = {lim sup}_{k} q_{k}$ is $\geq 0$ , so it is either $+ \infty$ , which is step 1.5, or real. In the real case step 2.1 with $ε = 1$ gives ${lim sup}_{k} r_{k} \leq β + 1$ , a real, so ${lim sup}_{k} r_{k} \neq + \infty$ ; if ${lim sup}_{k} r_{k} = - \infty$ it is $\leq β$ ; and otherwise it is a real $S$ , and $S > β$ would give, on choosing a natural $m \geq 1$ with $1 / m < S - β$ and applying step 2.1 with $ε = 1 / m$ , the impossibility $S \leq β + 1 / m < S$ . By totality ${lim sup}_{k} r_{k} \leq β$ .

step 2.1step 1.5step 1.1L2L11L12

3.2

Hence ${lim inf}_{k} q_{k} \leq {lim inf}_{k} r_{k}$ . By step 1.1 the element $α = {lim inf}_{k} q_{k}$ is $\geq 0$ , so it is $0$ , or a positive real, or $+ \infty$ . The first case is step 2.2. If $α$ is a positive real and ${lim inf}_{k} r_{k} < α$ , then ${lim inf}_{k} r_{k}$ lies between the reals $0$ and $α$ by step 1.1 and is therefore real, so [L13] supplies a real $c$ with ${lim inf}_{k} r_{k} < c < α$ , necessarily $c > 0$ ; step 2.3 then gives ${lim inf}_{k} r_{k} \geq c$ , contradicting $c > {lim inf}_{k} r_{k}$ , so ${lim inf}_{k} r_{k} \geq α$ by totality. If $α = + \infty$ , step 2.3 gives ${lim inf}_{k} r_{k} \geq c$ for every real $c$ with $c > 0$ , so ${lim inf}_{k} r_{k}$ is not $- \infty$ , and it is not a real $t$ either, since $t \geq 0$ by step 1.1 and then $c : = t + 1 > 0$ would give $t \geq t + 1$ ; hence ${lim inf}_{k} r_{k} = + \infty = α$ .

step 2.2step 2.3step 1.1L2L12L13

4.1

Combining the three links, ${lim inf}_{k} q_{k} \leq {lim inf}_{k} r_{k}$ by step 3.2, ${lim inf}_{k} r_{k} \leq {lim sup}_{k} r_{k}$ by [L4], and ${lim sup}_{k} r_{k} \leq {lim sup}_{k} q_{k}$ 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 $x_{k} \leq y_{k}$ eventually then $lim sup x_{k} \leq lim sup y_{k}$ and $lim inf x_{k} \leq lim inf y_{k}$ 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 $lim inf a_{k + 1} / a_{k} = 1 / 8$ , $lim sup a_{k + 1} / a_{k} = 2$ 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 ${lim sup}_{k} q_{k} < 1$ the chain forces ${lim sup}_{k} r_{k} < 1$ , so any conclusion drawn from the roots is available from the ratios; but ${lim sup}_{k} r_{k} < 1$ can hold with ${lim sup}_{k} q_{k} > 1$ , 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 $a_{k} \neq 0$ , and the roots must be defined, which needs $a_{k + 1} \geq 0$ ; 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 · two levels

89 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)
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>0ak​>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/akliminfak+1​/ak​≤liminfak1/k​≤limsupak1/k​≤limsupak+1​/ak​