TheoremStatement: Literature-sourcedProof: 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.

Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing

Statement

Let $(a_{k})_{k \geq 1}$ be a family of reals from the starting index $1$ (Series, partial sums, convergence and the sum, divergence, and the tail series), put

$ρ_{k} : = ∣ a_{k + 1} ∣^{1 / (k + 1)} (k \in N), ρ : = {lim sup}_{k} ρ_{k} \in \overline{R},$

and note that $ρ$ exists for every such family, with no hypothesis whatever (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence, 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 $\overline{R}$ ). Then:

if $ρ < 1$ then $\sum_{k \geq 1} ∣ a_{k} ∣$ converges, and hence $\sum_{k \geq 1} a_{k}$ converges as well;
if $ρ > 1$ then $\sum_{k \geq 1} a_{k}$ diverges;
if $ρ = 1$ neither conclusion follows: $\sum_{k \geq 1} 1 / k$ diverges, $\sum_{k \geq 1} 1 / k^{2}$ converges, and both have $ρ = 1$ .

The root family is shifted, and that is forced. The classical expression $∣ a_{n} ∣^{1 / n}$ is meaningful only for $n \geq 1$ , since $1 / 0$ is not a rational number (Rational powers $a^{r}$ of a positive base), while sequences here are functions on $N$ and $N$ contains $0$ . So the roots are written $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , which is $∣ a_{n} ∣^{1 / n}$ reindexed by $n = k + 1$ , exactly the convention of For $a_{k} > 0$ : $lim inf a_{k + 1} / a_{k} \leq lim inf a_{k}^{1 / k} \leq lim sup a_{k}^{1 / k} \leq lim sup a_{k + 1} / a_{k}$ . Every $ρ_{k}$ is defined, including where $a_{k + 1} = 0$ , by the supplementary clause of Rational powers $a^{r}$ of a positive base.

What claim 1 does and does not say. The comparison with a geometric series delivers convergence of the series of absolute values; that $\sum_{k \geq 1} a_{k}$ itself converges is a separate step, and it is supplied by If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges earlier on this page. Nothing here identifies the sum, and nothing here says anything about rearranging the series, which is taken up later in this track.

Facts & Assumptions

Given: A family $(a_{k})_{k \geq 1}$ of reals, the roots $ρ_{k} = ∣ a_{k + 1} ∣^{1 / (k + 1)}$ for $k \in N$ , the tail suprema $s_{n} = \sup {ρ_{k} : k \geq n}$ taken in $\overline{R}$ , and $ρ = {lim sup}_{k} ρ_{k} = \inf {s_{n} : n \in N}$ (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 extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[L1]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound there, and the extended order is total (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 $R$ , The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined). In particular $ρ \leq s_{n}$ for every $n$ , and $ρ_{k} \leq s_{n}$ for every $k \geq n$ ; a real $t$ with $ρ < t$ fails to be a lower bound of ${s_{n}}$ , and a real $u$ with $s_{n} > u$ fails to be an upper bound of ${ρ_{k} : k \geq n}$ .

[L2]

Both quantities exist for every sequence, bounded or not (The tail suprema of any real sequence are nonincreasing in $\overline{R}$ , so the limit superior exists for every sequence).

[L3]

Roots and powers: for $x \geq 0$ and natural $n \geq 1$ , $x^{1 / n} \geq 0$ and $(x^{1 / n})^{n} = x$ ; on the nonnegatives $y \mapsto y^{n}$ is strictly increasing for $n \geq 1$ ; and $1^{n} = 1$ (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, Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L4]

Absolute value: $∣ x ∣ \geq 0$ for every real $x$ (Basic properties of the absolute value).

[L5]

The geometric series $\sum_{j \geq 0} t^{j}$ converges when $∣ t ∣ < 1$ , and a series converges if and only if each of its tail series converges (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L6]

Direct comparison, in the form for families from a general starting index: if $0 \leq x_{k} \leq y_{k}$ from some index on and $\sum y_{k}$ converges then $\sum x_{k}$ converges (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 $\sum b_{k}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

If a series converges then its terms tend to $0$ ; contrapositively, terms not tending to $0$ force divergence (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L8]

$1 \leq n^{1 / n}$ for every natural $n \geq 1$ , and the sequence $(k + 1)^{1 / (k + 1)}$ converges to $1$ ( $n^{1 / n} \to 1$ ); a sequence converging to a real $c$ has $lim sup = lim inf = c$ (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 $\pm \infty$ ); products and quotients of convergent sequences converge, the quotient requiring nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients).

[L9]

Laws of rational exponents on a positive base: $(a^{r})^{s} = a^{r s}$ , $a^{- r} = 1 / a^{r}$ and $a^{r} > 0$ ; and for rational $t > 0$ , $a > 1$ implies $a^{t} > 1$ (Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L10]

For rational $p > 0$ , $\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[L11]

If $\sum ∣ x_{j} ∣$ converges then $\sum x_{j}$ converges; for a family from the starting index $1$ this is the same statement applied to the shifted sequence $j \mapsto a_{j + 1}$ , whose series is $\sum_{k \geq 1} a_{k}$ and whose absolute-value series is $\sum_{k \geq 1} ∣ a_{k} ∣$ (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · cases

1.1

Assume $ρ < 1$ .

assume-case lt

1.2

Assume instead $ρ > 1$ .

assume-case gt

1.3

Assume instead $ρ = 1$ .

assume-case one

2.1

Every $ρ_{k}$ is a nonnegative real, so each $s_{n} \geq ρ_{n} \geq 0$ and hence $0$ is a lower bound of ${s_{n}}$ , giving $ρ \geq 0$ ; combined with the case hypothesis $ρ < 1$ this puts $ρ$ strictly between the reals $0$ and $1$ , so $ρ$ is a real number.

step 1.1L1L2L3L4

2.2

In the case $ρ > 1$ , the value $ρ$ is a lower bound of ${s_{n}}$ , so $s_{n} \geq ρ > 1$ for every $n \in N$ .

step 1.2L1

2.3

In the case $ρ = 1$ , take first $b_{k} : = 1 / k$ for $k \geq 1$ . Its root family is $(1 / (k + 1))^{1 / (k + 1)} = 1 / (k + 1)^{1 / (k + 1)}$ , and since $(k + 1)^{1 / (k + 1)} \to 1$ with every term at least $1$ , the quotient rule gives convergence to $1$ , so the limit superior of the root family is $1$ ; and $\sum_{k \geq 1} 1 / k$ diverges, being the case $p = 1$ .

step 1.3L8L9L10

2.4

In the case $ρ = 1$ , take next $c_{k} : = 1 / k^{2}$ for $k \geq 1$ . Its root family is $((k + 1)^{- 2})^{1 / (k + 1)} = ((k + 1)^{1 / (k + 1)})^{- 2}$ , which converges to $1^{- 2} = 1$ by the product and quotient rules, so again the limit superior of the root family is $1$ ; and $\sum_{k \geq 1} 1 / k^{2}$ converges, being the case $p = 2$ .

step 1.3L8L9L10

3.1

In the case $ρ < 1$ put $t : = (ρ + 1) / 2$ , a real number with $0 \leq ρ < t < 1$ ; since $t$ is not a lower bound of ${s_{n}}$ there is $N \in N$ with $s_{N} < t$ .

step 2.1L1choose

3.2

In the case $ρ > 1$ , for each $n$ the real $1$ is not an upper bound of ${ρ_{k} : k \geq n}$ , so there is $k \geq n$ with $ρ_{k} > 1$ .

step 2.2L1

3.3

So at $ρ = 1$ one family gives a divergent series and another a convergent one, and neither of the two conclusions can be drawn, which is claim 3.

step 2.3step 2.4

4.1

In the case $ρ < 1$ , for every $k \geq N$ we have $ρ_{k} \leq s_{N} < t$ , and raising both nonnegative sides to the power $k + 1 \geq 1$ gives $∣ a_{k + 1} ∣ = (ρ_{k})^{k + 1} < t^{k + 1}$ .

step 3.1L1L3

4.2

In the case $ρ > 1$ , whenever $ρ_{k} > 1$ we get $∣ a_{k + 1} ∣ = (ρ_{k})^{k + 1} > 1^{k + 1} = 1$ ; so by step 3.2 there are indices $k \geq n$ with $∣ a_{k + 1} ∣ > 1$ for every $n$ .

step 3.2L3

4.3

In the case $ρ < 1$ : since $0 < t < 1$ the geometric series $\sum_{j \geq 0} t^{j}$ converges, hence so does its first tail series $\sum_{m \geq 1} t^{m}$ .

step 3.1L5

5.1

In the case $ρ < 1$ : putting $x_{j} : = ∣ a_{j + 1} ∣$ and $y_{j} : = t^{j + 1}$ for $j \in N$ , step 4.1 gives $0 \leq x_{j} \leq y_{j}$ for all $j \geq N$ , and $\sum_{j} y_{j}$ is the convergent series of step 4.3; so $\sum_{k \geq 1} ∣ a_{k} ∣$ converges.

step 4.1step 4.3L4L6

5.2

In the case $ρ > 1$ : the sequence $j \mapsto a_{j + 1}$ does not converge to $0$ , because with the rational tolerance $1$ no index $K$ satisfies $∣ a_{k + 1} ∣ < 1$ for all $k \geq K$ ; hence $\sum_{k \geq 1} a_{k}$ diverges, which is claim 2.

step 4.2L7

6.1

In the case $ρ < 1$ : the series $\sum_{k \geq 1} ∣ a_{k} ∣$ having been shown to converge, the sequence $j \mapsto a_{j + 1}$ has a convergent absolute-value series, so $\sum_{k \geq 1} a_{k}$ converges as well; together with the convergence of $\sum_{k \geq 1} ∣ a_{k} ∣$ that is claim 1.

step 5.1L11

7.1

The three cases $ρ < 1$ , $ρ > 1$ and $ρ = 1$ exhaust $\overline{R}$ , the extended order being total, so the three claims together cover every family.

step 6.1step 5.2step 3.3L1cases-exhaustive∎

Remarks

The test reads only the tail suprema, and that is why it never needs the roots to converge. Claim 1 uses a single index $N$ beyond which all roots sit below a fixed $t < 1$ ; claim 2 uses only that roots above $1$ occur arbitrarily late. Neither argument asks whether $(ρ_{k})$ has a limit, which is exactly the advantage of $lim sup$ over $\lim$ here.
Claim 2 is proved through the term test, not through a comparison. What the hypothesis delivers is infinitely many terms of absolute value greater than $1$ , which already forbids the terms from tending to $0$ . No estimate on the partial sums is needed, and none is available, the terms having no sign.
The witnesses in claim 3 are chosen so that both root computations reduce to the single standard limit $n^{1 / n} \to 1$ . The companion page carries the same phenomenon with the exponents $- 1 / 2$ and $- 2$ , where the divergent witness is not the harmonic series.

Depends on

Used by

Dependency tree · two levels

87 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)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.33) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

Root test: lim sup⁡∣ak∣1/k<1limsup∣ak​∣1/k<1 gives absolute convergence and hence convergence, >1>1 gives divergence, and =1=1 decides nothing