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False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak for every positive sequence

Statement

False claim: for every sequence (ak) of reals with ak>0 for all k,

lim sup⁡kak+11/(k+1)  =  lim sup⁡kak+1ak,

that is, the limit superior of the root sequence equals the limit superior of the ratio sequence. (The root family is written with the shift of For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak, since ak1/k is undefined at k=0; classically the claim reads lim sup⁡nan1/n=lim sup⁡nan+1/an.)

What is true is the chain lim inf⁡kak+1ak  ≤  lim inf⁡kak+11/(k+1)  ≤  lim sup⁡kak+11/(k+1)  ≤  lim sup⁡kak+1ak of For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak. The claim above collapses its right-hand inequality to an equality, and that fails: the roots can converge while the ratios oscillate. This is exactly why a root criterion decides cases that a ratio criterion cannot.

The witness is ak=2−k+(−1)k. The computation below establishes all four quantities for it, namely lim inf⁡kak+1ak=18,lim sup⁡kak+1ak=2,lim⁡kak+11/(k+1)=12, and it is recorded as a named example on the companion page.

Facts & Assumptions

Given: The alternating sequence (sk) and the index maps e,o of The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1; the sequence tk defined to be 2 when sk=1 and 1/2 when sk=−1; the sequence ak:=2−ktk; the ratios qk:=ak+1/ak and the roots rk:=ak+11/(k+1).

[L1]

The alternating sequence: s0=1, sk+1=−sk, ∣sk∣=1 for every k, sej=1 and soj=−1, and e, o are strictly increasing (The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1); a strictly increasing index map satisfies nj≥j (A strictly increasing index map satisfies nk≥k).

[L4]

Powers and roots of positive reals: integer powers with 2m2m′=2m+m′ and 2−m=1/2m; (xy)1/n=x1/ny1/n; the integer power is the rational power at an integer exponent, so (2−n)1/n=2−n/n=2−1; roots of positive reals are positive; and 0<x≤y implies x1/n≤y1/n (Integer powers am, Laws of integer exponents, Rational powers ar of a positive base, Laws of rational exponents, Monotonicity of r↦ar and of a↦ar, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a).

[L5]

For every real b>0 the sequence b1/(k+1) converges to 1 (For every a>0, a1/n→1).

[L9]

The refuted claim: for every sequence of positive reals, lim sup⁡kak+11/(k+1)=lim sup⁡kak+1/ak.

Refutation

technique · direct
1.1

Each sk is 1 or −1 because ∣sk∣=1, so tk is well defined, with tk∈{2,1/2} and tk>0; hence ak=2−ktk>0 for every k, and (ak) is a sequence of positive reals to which the claim applies. This is the sequence usually written ak=2−k+(−1)k.

givenL1L4L7L9
1.2

Since sk+1=−sk and 1≠−1, exactly one of the two situations "sk=1 and sk+1=−1" and "sk=−1 and sk+1=1" occurs at each index k. In the first, tk=2 and tk+1=1/2; in the second, tk=1/2 and tk+1=2.

givenL1L7
1.3

For every n both values of s occur at an index ≥n: sen=1 with en≥n and son=−1 with on≥n.

givenL1
2.1

The ratios are qk=ak+1/ak=(2−(k+1)tk+1)/(2−ktk)=2−1tk+1/tk, which by step 1.2 equals 2−1(1/2)/2=1/8 when sk=1 and 2−1⋅2/(1/2)=2 when sk=−1.

step 1.1step 1.2L4L7algebra
2.2

The roots are rk=(2−(k+1)tk+1)1/(k+1)=(2−(k+1))1/(k+1)tk+11/(k+1)=2−1tk+11/(k+1).

step 1.1L4
2.3

Since 1/2≤tk+1≤2 for every k and x↦x1/(k+1) is nondecreasing on the positive reals, (1/2)1/(k+1)≤tk+11/(k+1)≤21/(k+1); both bounding sequences converge to 1 by [L5], so the squeeze theorem gives tk+11/(k+1)→1.

step 1.1L4L5L6L7
3.1

By steps 1.2 and 1.3 the tail range of (qk) at every index n is exactly {1/8,2}: those are the only values, and each occurs at some index ≥n. Its least upper bound in R‾ is 2 and its greatest lower bound is 1/8, since 1/8<2 and both belong to the set; hence lim sup⁡kqk is the greatest lower bound of {2}, namely 2, and lim inf⁡kqk is the least upper bound of {1/8}, namely 1/8.

step 2.1step 1.3L2L7
3.2

By steps 2.2 and 2.3 and the scalar rule, rk=2−1tk+11/(k+1)→2−1⋅1=1/2, so lim sup⁡krk=lim inf⁡krk=1/2.

step 2.2step 2.3L3L6
4.1

For this sequence the claim asserts lim sup⁡krk=lim sup⁡kqk, that is 1/2=2; but 1/2<1<2, so the two are different and the claim fails.

step 3.1step 3.2L7L9
5.1

The claim is therefore false. The true chain [L8] reads here 1/8≤1/2≤1/2≤2, so both outer inequalities are strict for this witness while the middle one is an equality.

step 4.1step 3.1step 3.2L7L8L9∎

Remarks

Depends on

Used by

Dependency tree · two levels

84 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