Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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A positive sequence making all three inequalities of the ratio-to-root chain strict

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 and define

ak:=2−k  when sk=1,ak:=3−k  when sk=−1.

This interleaves the two geometric sequences 2−k and 3−k, taking the first at even indices and the second at odd ones. With qk:=ak+1/ak and rk:=ak+11/(k+1) as in For ak>0: lim inf⁡ak+1/ak≤lim inf⁡ak1/k≤lim sup⁡ak1/k≤lim sup⁡ak+1/ak,

lim inf⁡kqk=0,lim inf⁡krk=13,lim sup⁡krk=12,lim sup⁡kqk=+∞,

so the chain of that theorem reads

0  <  13  <  12  <  +∞

with all three inequalities strict.

Where each comparison lives. The first two, 0<1/3 and 1/3<1/2, are comparisons of real numbers and hold in R; they hold in R‾ as well only because the extended order restricts on R to the order of R (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined). The third, 1/2<+∞, is not a comparison in R at all: +∞ is not a real number, and the inequality is the instance of "every real is below the greatest element" in R‾. So the outer two values of the chain are of different kinds here, and only the extended line can hold all four at once.

Facts & Assumptions

Given: The alternating sequence (sk) with index maps e,o (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 ak defined above; the ratios qk=ak+1/ak; and the roots rk=ak+11/(k+1).

[L1]

The alternating sequence: ∣sk∣=1, sk+1=−sk, sej=1, soj=−1, with e, o strictly increasing, so ej≥j and oj≥j; also o0=σ(0)≥1 (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 nk≥k).

[L3]

The order on R‾ is total, +∞ is greatest, every real is <+∞, and the order restricts on R to the order of R (The extended real line R‾=R∪{−∞,+∞}, its order, and the arithmetic that is left undefined).

[L4]

Powers: 2−k=(1/2)k and 3−k=(1/3)k; xmxm′=xm+m′ and (xy)m=xmym for integer exponents and nonzero bases; xm>0 for x>0; (x−n)1/n=x−1 for x>0 and n≥1 (Integer powers am, Laws of integer exponents, Rational powers ar of a positive base, Laws of rational exponents, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a).

[L5]

Geometric sequences: ∣ρ∣<1 implies ρk→0, and ∣ρ∣>1 implies ∣ρ∣k→+∞ (For ∣r∣<1 the sequence rk is null, and for ∣r∣>1 the sequence ∣r∣k diverges to +∞, Limits and Cauchy sequences of reals, Divergence to +∞ and to −∞).

[L7]

The order on N is total, so any two indices have a common upper bound (Order on the natural numbers, ≤ is a linear order on N).

[L8]

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).

Verification

technique · direct
1.1

Each sk is 1 or −1, so ak is well defined, and ak>0 for every k since positive powers of positive bases are positive.

givenL1L4L6
1.2

For every n∈N there are indices k,k′≥n with sk=1 and sk′=−1, namely k=en and k′=on; and there are indices l,l′≥n with sl+1=1 and sl′+1=−1, namely l=ej−1 and l′=oj−1 for any j≥n+1, these being natural numbers because ej≥j≥1 and oj≥j≥1, and satisfying l≥j−1≥n and l′≥j−1≥n.

givenL1L7
1.3

Since sk+1=−sk, the ratios are qk=3−(k+1)/2−k=3−1(2/3)k when sk=1, and qk=2−(k+1)/3−k=2−1(3/2)k when sk=−1; in both cases qk>0.

givenL1L4L6
1.4

Likewise the roots are rk=(2−(k+1))1/(k+1)=2−1 when sk+1=1, and rk=(3−(k+1))1/(k+1)=3−1 when sk+1=−1.

givenL1L4
2.1

By steps 1.2 and 1.4 the tail range of (rk) at every index n is exactly {1/2,1/3}, whose least upper bound is 1/2 and greatest lower bound 1/3, since 1/3<1/2 and both belong to the set. Hence lim sup⁡krk=1/2 and lim inf⁡krk=1/3.

step 1.2step 1.4L2L3L6
2.2

lim sup⁡kqk=+∞. Fix n and a real M. Since ∣3/2∣>1, the sequence (3/2)k diverges to +∞, so there is K with 2−1(3/2)k>M for all k≥K; taking j at least as large as both n and K and putting k:=oj, we get k≥j≥n and sk=−1, hence qk=2−1(3/2)k>M. So no real bounds the tail range of (qk) above, its least upper bound in R‾ is +∞ for every n, and lim sup⁡kqk is the greatest lower bound of {+∞}, namely +∞.

step 1.2step 1.3L2L3L5L6L7
2.3

lim inf⁡kqk=0. Fix n. All qk are positive, so 0 is a lower bound of the tail range. If ℓ>0 were a lower bound, then, since ∣2/3∣<1 gives (2/3)k→0 and hence 3−1(2/3)k<ℓ for all k≥K for some K, taking j at least as large as both n and K and putting k:=ej would give k≥j≥n, sk=1 and qk=3−1(2/3)k<ℓ, contradicting that ℓ is a lower bound. So every lower bound is ≤0 and the greatest lower bound of each tail range is 0; hence lim inf⁡kqk is the least upper bound of {0}, namely 0.

step 1.2step 1.3L2L3L5L6L7
3.1

Collecting the four values, the chain [L8] reads 0≤1/3≤1/2≤+∞, and each inequality is strict: 0<1/3 and 1/3<1/2 hold in R and therefore in R‾, while 1/2<+∞ holds because +∞ is the greatest element of R‾ and is distinct from every real. So no two of the four quantities coincide.

step 2.1step 2.2step 2.3L3L6L8∎

Remarks

Depends on

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