Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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∑k−1/2 diverges and ∑k−2 converges, and both have root limit exactly 1

Statement refuted

Refuted claim: the value lim sup⁡k∣ak+1∣1/(k+1)=1 determines the behaviour of ∑k≥1ak; 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

ak:=k−1/2,bk:=k−2(k≥1),

rational powers of the canonical naturals (Rational powers ar of a positive base). Both have root quantity exactly 1, while ∑k≥1ak diverges and ∑k≥1bk converges (For rational p>0, ∑1/kp converges iff p>1, at p=1/2 and p=2).

So the third clause of Root test: lim sup⁡∣ak∣1/k<1 gives absolute convergence and hence convergence, >1 gives divergence, and =1 decides nothing is not a gap in the proof: at lim sup⁡=1 nothing whatever follows, and the two witnesses here are on opposite sides.

Facts & Assumptions

Given: The families ak:=ι(k)−1/2 and bk:=ι(k)−2 for naturals k≥1; the sequence uj:=ι(j+1)1/(j+1), j∈N; and the root families αj:=aj+11/(j+1), βj:=bj+11/(j+1) (Rational powers ar of a positive base, Canonical naturals are positive and strictly increasing).

[L1]

1≤n1/n for every natural n≥1, and uj=(j+1)1/(j+1)→1 (n1/n→1).

[L2]

Laws of rational exponents on a positive base: (xr)s=xrs, x−r=1/xr, xr>0, and 1r=1 (Laws of rational exponents, Rational powers ar of a positive base).

[L3]

Monotonicity of rational powers: for rational t>0, 0<x≤y implies xt≤yt; and for x≥1 and rationals r<s, xr≤xs (Monotonicity of r↦ar and of a↦ar).

[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).

[L7]

The canonical naturals are positive with ι(k)≥1 for k≥1; reciprocation reverses the order on the positives; and ∣x∣=x for x≥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≥1 we have ι(k)≥1>0, so ak and bk are positive and equal to their own absolute values; and uj≥1>0 for every j.

givenL1L2L7
1.2

The root family of (bk) is βj=(ι(j+1)−2)1/(j+1)=(ι(j+1)1/(j+1))−2=1/uj2.

givenL2
1.3

The series ∑k≥1k−1/2 is the p-series at p=1/2, and 1/2>1 is false, so it diverges.

givenL6
1.4

The series ∑k≥1k−2 is the p-series at p=2, and 2>1, so it converges.

givenL6
2.1

Since ι(j+1)≥1 and −1<−1/2<0, we have ι(j+1)−1≤ι(j+1)−1/2≤ι(j+1)0=1.

step 1.1L3L2
2.2

Since uj→1, the product rule gives uj2→1, and the quotient rule then gives βj→1; so lim sup⁡jβj=1.

step 1.2L1L4L5
3.1

The root family of (ak) is αj=(ι(j+1)−1/2)1/(j+1)=ι(j+1)−1/(2(j+1)), and applying the same exponent to the two bounds of step 2.1 gives 1/uj=(ι(j+1)−1)1/(j+1)≤αj≤11/(j+1)=1.

step 2.1L2L3
4.1

Since uj→1 with uj≥1>0, the quotient rule gives 1/uj→1; so αj→1 by the squeeze theorem, and therefore lim sup⁡jαj=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 uj−p, which tends to 1 because uj 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

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Dependency tree · two levels

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Sources