CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$\sum k^{-1/2}$ diverges and $\sum k^{-2}$ converges, and both have root limit exactly $1$

Statement refuted

Refuted claim: the value $\limsup_k |a_{k+1}|^{1/(k+1)} = 1$ determines the behaviour of $\sum_{k \ge 1} a_k$ ; that is, any two series with root quantity equal to $1$ either both converge or both diverge.

The claim is refuted by the two families

$a_k := k^{-1/2}, \qquad b_k := k^{-2} \qquad (k \ge 1),$

rational powers of the canonical naturals (Rational powers $a^r$ of a positive base). Both have root quantity exactly $1$ , while $\sum_{k \ge 1} a_k$ diverges and $\sum_{k \ge 1} b_k$ converges (For rational $p > 0$ , $\sum 1/k^p$ converges iff $p > 1$ , at $p = 1/2$ and $p = 2$ ).

So the third clause of Root test: $\limsup |a_k|^{1/k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing is not a gap in the proof: at $\limsup = 1$ nothing whatever follows, and the two witnesses here are on opposite sides.

Facts & Assumptions

Given: The families $a_k := \iota(k)^{-1/2}$ and $b_k := \iota(k)^{-2}$ for naturals $k \ge 1$ ; the sequence $u_j := \iota(j+1)^{1/(j+1)}$ , $j \in \mathbb{N}$ ; and the root families $\alpha_j := a_{j+1}^{1/(j+1)}$ , $\beta_j := b_{j+1}^{1/(j+1)}$ (Rational powers $a^r$ of a positive base, Canonical naturals are positive and strictly increasing).

[L1]

$1 \le n^{1/n}$ for every natural $n \ge 1$ , and $u_j = (j+1)^{1/(j+1)} \to 1$ ( $n^{1/n} \to 1$ ).

[L2]

Laws of rational exponents on a positive base: $(x^{r})^{s} = x^{rs}$ , $x^{-r} = 1/x^{r}$ , $x^{r} > 0$ , and $1^{r} = 1$ (Laws of rational exponents, Rational powers $a^r$ of a positive base).

[L3]

Monotonicity of rational powers: for rational $t > 0$ , $0 < x \le y$ implies $x^{t} \le y^{t}$ ; and for $x \ge 1$ and rationals $r < s$ , $x^{r} \le x^{s}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L4]

The squeeze theorem, and the product and quotient rules for limits, the quotient requiring a nonzero limit and nonzero denominators (The squeeze theorem, Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L5]

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 $\pm\infty$ , 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 $\overline{\mathbb{R}}$ ).

[L6]

$\sum_{k \ge 1} 1/k^{p}$ converges if and only if $p > 1$ (For rational $p > 0$ , $\sum 1/k^p$ converges iff $p > 1$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

The canonical naturals are positive with $\iota(k) \ge 1$ for $k \ge 1$ ; reciprocation reverses the order on the positives; and $|x| = x$ for $x \ge 0$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value).

Counterexample

technique · direct

1.1

For $k \ge 1$ we have $\iota(k) \ge 1 > 0$ , so $a_k$ and $b_k$ are positive and equal to their own absolute values; and $u_j \ge 1 > 0$ for every $j$ .

givenL1L2L7

1.2

The root family of $(b_k)$ is $\beta_j = \big(\iota(j+1)^{-2}\big)^{1/(j+1)} = \big(\iota(j+1)^{1/(j+1)}\big)^{-2} = 1/u_j^{2}$ .

givenL2

1.3

The series $\sum_{k \ge 1} k^{-1/2}$ is the $p$ -series at $p = 1/2$ , and $1/2 > 1$ is false, so it diverges.

givenL6

1.4

The series $\sum_{k \ge 1} k^{-2}$ is the $p$ -series at $p = 2$ , and $2 > 1$ , so it converges.

givenL6

2.1

Since $\iota(j+1) \ge 1$ and $-1 < -1/2 < 0$ , we have $\iota(j+1)^{-1} \le \iota(j+1)^{-1/2} \le \iota(j+1)^{0} = 1$ .

step 1.1L3L2

2.2

Since $u_j \to 1$ , the product rule gives $u_j^{2} \to 1$ , and the quotient rule then gives $\beta_j \to 1$ ; so $\limsup_j \beta_j = 1$ .

step 1.2L1L4L5

3.1

The root family of $(a_k)$ is $\alpha_j = \big(\iota(j+1)^{-1/2}\big)^{1/(j+1)} = \iota(j+1)^{-1/(2(j+1))}$ , and applying the same exponent to the two bounds of step 2.1 gives $1/u_j = \big(\iota(j+1)^{-1}\big)^{1/(j+1)} \le \alpha_j \le 1^{1/(j+1)} = 1$ .

step 2.1L2L3

4.1

Since $u_j \to 1$ with $u_j \ge 1 > 0$ , the quotient rule gives $1/u_j \to 1$ ; so $\alpha_j \to 1$ by the squeeze theorem, and therefore $\limsup_j \alpha_j = 1$ .

step 3.1L1L4L5

5.1

Both families have root quantity exactly $1$ , yet one series diverges and the other converges; the claim is refuted, and the third clause of the root test is confirmed as unavoidable.

step 4.1step 2.2step 1.3step 1.4∎

Remarks

Every $p$ -series has root quantity $1$ . The computation in step 1.2 generalises verbatim: for rational $p > 0$ the root family of $k^{-p}$ is $u_j^{-p}$ , which tends to $1$ because $u_j$ does. So the root test is silent on the entire $p$ -series family, which is precisely the family the condensation test settles.
The root test and the ratio test are silent on the same family. The ratios of $k^{-p}$ also tend to $1$ , so neither test separates $p = 1/2$ from $p = 2$ . What does separate them is Raabe's test, whose expression reads the rate at which the ratios approach $1$ ; the companion example on this page carries the case $p = 2$ .
Why the two exponents are $-1/2$ and $-2$ rather than $-1$ and $-2$ . Taking the divergent witness with a fractional exponent makes the point that the failure is not about the harmonic series in particular: the root quantity is blind to the exponent altogether, and any pair straddling $p = 1$ would do.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Root test (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
Convergence tests (Wikipedia) (standard reference, not scraped)

∑k−1/2\sum k^{-1/2}∑k−1/2 diverges and ∑k−2\sum k^{-2}∑k−2 converges, and both have root limit exactly 111