Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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ak=2−k+(−1)k has lim inf⁡ak+1/ak=1/8, lim sup⁡ak+1/ak=2 and lim⁡ak1/k=1/2

Example

Let (sk) be the alternating sequence 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, let tk:=2 when sk=1 and tk:=1/2 when sk=−1, and put

ak  :=  2−ktk(k∈N),

the sequence usually written ak=2−k+(−1)k. Writing qk:=ak+1/ak for the ratios and rk:=ak+11/(k+1) for the roots, which is an1/n reindexed by n=k+1 as For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak requires,

lim inf⁡kqk=18,lim sup⁡kqk=2,lim⁡krk=12,

so also lim inf⁡krk=lim sup⁡krk=1/2. In addition ak→0.

The point. The ratios oscillate across 1, taking the values 1/8 and 2 alternately, so no statement of the form "the ratios are eventually below some λ<1" is available; the roots, by contrast, converge to 1/2<1. Any criterion reading the ratios alone is silent here, and one reading the roots is not. That is the concrete form of the dominance recorded in For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak.

Facts & Assumptions

Given: The alternating sequence (sk), the auxiliary tk∈{2,1/2}, the sequence ak=2−ktk, the ratios qk=ak+1/ak and the roots rk=ak+11/(k+1), all as in FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak for every positive sequence.

[L1]

For this sequence: every ak is positive, qk∈{1/8,2} with both values occurring at arbitrarily large indices, lim inf⁡kqk=1/8, lim sup⁡kqk=2, and rk→1/2, so lim inf⁡krk=lim sup⁡krk=1/2 (FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak for every positive sequence).

[L2]

The chain lim inf⁡kqk≤lim inf⁡krk≤lim sup⁡krk≤lim sup⁡kqk (For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak).

[L4]

Powers: 2−k=(1/2)k and tk≤2, so ak≤2⋅(1/2)k; and ak>0 (Integer powers am, Laws of integer exponents, Rational powers ar of a positive base, Sign rules for products and monotonicity of multiplication).

Verification

technique · direct
1.1

The three values in the display are exactly what [L1] records for this sequence, together with lim inf⁡krk=lim sup⁡krk=1/2, which follows from the convergence of (rk) to 1/2.

givenL1L3
1.2

The sequence is null: 0<ak=2−ktk≤2⋅(1/2)k for every k, and (1/2)k→0 because ∣1/2∣<1, so 2⋅(1/2)k→0 and the squeeze gives ak→0.

givenL4L5L6
2.1

The ratio quantities differ from one another and from the root quantities: 1/8<1/2<2, so lim inf⁡kqk<lim inf⁡krk=lim sup⁡krk<lim sup⁡kqk. In particular the chain [L2] holds here with both outer inequalities strict and the middle one an equality, and the ratios do not determine the roots.

step 1.1L1L2L6
3.1

So (ak) is a positive null sequence whose root sequence converges to 1/2<1 while its ratio sequence has lim sup⁡kqk=2>1 and lim inf⁡kqk=1/8<1, that is, the ratios oscillate across 1 while the roots settle strictly below it.

step 1.2step 2.1L1L6∎

Remarks

  • Where the numbers come from. The exponent −k+(−1)k changes by −1+(−1)k+1−(−1)k=−1∓2 from one index to the next, giving ratios 2−3=1/8 and 21=2; the root divides the exponent by the index, so the bounded oscillation contributes 2±1/(k+1)→1 and only the linear part −k survives, giving 2−1=1/2. The full computation is in FALSE: lim sup⁡ak1/k=lim sup⁡ak+1/ak for every positive sequence.

  • The same sequence reappears for series. With these ak the series ∑kak converges, and the root criterion sees it while the ratio criterion does not. That use belongs to the series page and is not made here.

  • Strictness of the middle inequality needs a different witness. Here lim inf⁡krk=lim sup⁡krk, since the roots converge. A sequence making all three inequalities of the chain strict is A positive sequence making all three inequalities of the ratio-to-root chain strict.

Depends on

Used by

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Sources