TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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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.

Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing

Statement

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

$\rho_k \;:=\; |a_{k+1}|^{1/(k+1)} \qquad (k \in \mathbb{N}), \qquad \rho \;:=\; \limsup_{k} \rho_k \;\in\; \overline{\mathbb{R}} ,$

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

if $\rho < 1$ then $\sum_{k \ge 1} |a_k|$ converges, and hence $\sum_{k \ge 1} a_k$ converges as well;
if $\rho > 1$ then $\sum_{k \ge 1} a_k$ diverges;
if $\rho = 1$ neither conclusion follows: $\sum_{k \ge 1} 1/k$ diverges, $\sum_{k \ge 1} 1/k^{2}$ converges, and both have $\rho = 1$ .

The root family is shifted, and that is forced. The classical expression $|a_n|^{1/n}$ is meaningful only for $n \ge 1$ , since $1/0$ is not a rational number (Rational powers $a^r$ of a positive base), while sequences here are functions on $\mathbb{N}$ and $\mathbb{N}$ contains $0$ . So the roots are written $\rho_k = |a_{k+1}|^{1/(k+1)}$ , which is $|a_n|^{1/n}$ reindexed by $n = k+1$ , exactly the convention of For $a_k > 0$ : $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ . Every $\rho_k$ 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 $\sum_{k \ge 1} a_k$ 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 $(a_k)_{k \ge 1}$ of reals, the roots $\rho_k = |a_{k+1}|^{1/(k+1)}$ for $k \in \mathbb{N}$ , the tail suprema $s_n = \sup\{\rho_k : k \ge n\}$ taken in $\overline{\mathbb{R}}$ , and $\rho = \limsup_k \rho_k = \inf\{s_n : n \in \mathbb{N}\}$ (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 extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$ , its order, and the arithmetic that is left undefined).

[L1]

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

[L2]

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

[L3]

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

[L4]

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

[L5]

The geometric series $\sum_{j \ge 0} t^{j}$ converges when $|t| < 1$ , and a series converges if and only if each of its tail series converges (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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 $0 \le x_k \le y_k$ from some index on and $\sum y_k$ converges then $\sum x_k$ converges (If $0 \le a_k \le b_k$ 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]

$1 \le n^{1/n}$ for every natural $n \ge 1$ , and the sequence $(k+1)^{1/(k+1)}$ converges to $1$ ( $n^{1/n} \to 1$ ); a sequence converging to a real $c$ has $\limsup = \liminf = c$ (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 $\pm\infty$ ); 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: $(a^{r})^{s} = a^{rs}$ , $a^{-r} = 1/a^{r}$ 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 $p > 0$ , $\sum_{k \ge 1} 1/k^{p}$ 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 $\sum_{k \ge 1} a_k$ and whose absolute-value series is $\sum_{k \ge 1} |a_k|$ (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 $\rho < 1$ .

assume-case lt

1.2

Assume instead $\rho > 1$ .

assume-case gt

1.3

Assume instead $\rho = 1$ .

assume-case one

2.1

Every $\rho_k$ is a nonnegative real, so each $s_n \ge \rho_n \ge 0$ and hence $0$ is a lower bound of $\{s_n\}$ , giving $\rho \ge 0$ ; combined with the case hypothesis $\rho < 1$ this puts $\rho$ strictly between the reals $0$ and $1$ , so $\rho$ is a real number.

step 1.1L1L2L3L4

2.2

In the case $\rho > 1$ , the value $\rho$ is a lower bound of $\{s_n\}$ , so $s_n \ge \rho > 1$ for every $n \in \mathbb{N}$ .

step 1.2L1

2.3

In the case $\rho = 1$ , take first $b_k := 1/k$ for $k \ge 1$ . Its root family is $\big(1/(k+1)\big)^{1/(k+1)} = 1 / (k+1)^{1/(k+1)}$ , 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 $\sum_{k \ge 1} 1/k$ diverges, being the case $p = 1$ .

step 1.3L8L9L10

2.4

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

step 1.3L8L9L10

3.1

In the case $\rho < 1$ put $t := (\rho + 1)/2$ , a real number with $0 \le \rho < t < 1$ ; since $t$ is not a lower bound of $\{s_n\}$ there is $N \in \mathbb{N}$ with $s_N < t$ .

step 2.1L1choose

3.2

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

step 2.2L1

3.3

So at $\rho = 1$ 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 $\rho < 1$ , for every $k \ge N$ we have $\rho_k \le s_N < t$ , and raising both nonnegative sides to the power $k+1 \ge 1$ gives $|a_{k+1}| = (\rho_k)^{k+1} < t^{\,k+1}$ .

step 3.1L1L3

4.2

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

step 3.2L3

4.3

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

step 3.1L5

5.1

In the case $\rho < 1$ : putting $x_j := |a_{j+1}|$ and $y_j := t^{\,j+1}$ for $j \in \mathbb{N}$ , step 4.1 gives $0 \le x_j \le y_j$ for all $j \ge N$ , and $\sum_j y_j$ is the convergent series of step 4.3; so $\sum_{k \ge 1} |a_k|$ converges.

step 4.1step 4.3L4L6

5.2

In the case $\rho > 1$ : 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 $k \ge K$ ; hence $\sum_{k \ge 1} a_k$ diverges, which is claim 2.

step 4.2L7

6.1

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

step 5.1L11

7.1

The three cases $\rho < 1$ , $\rho > 1$ and $\rho = 1$ exhaust $\overline{\mathbb{R}}$ , 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 $(\rho_k)$ has a limit, which is exactly the advantage of $\limsup$ 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.

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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<1\limsup |a_k|^{1/k} < 1limsup∣ak​∣1/k<1 gives absolute convergence and hence convergence, >1> 1>1 gives divergence, and =1= 1=1 decides nothing