Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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Raabe is Kummer with ζk=k+1\zeta_k = k+1: for positive terms, lim inf(k+1)(ak/ak+11)>1\liminf\, (k+1)(a_k/a_{k+1} - 1) > 1 gives convergence and lim sup<1\limsup < 1 gives divergence

Statement

Let (ak)(a_k) be a sequence of reals with ak>0a_k > 0 for every kNk \in \mathbb{N}. Write k+1k+1 for the canonical natural ι(k+1)R\iota(k+1) \in \mathbb{R}, which is positive (Canonical naturals are positive and strictly increasing), take the weights ζk:=k+1\zeta_k := k+1 in Kummer: for positive terms aka_k and weights ζk>0\zeta_k > 0, lim inf(ζkak/ak+1ζk+1)>0\liminf(\zeta_k a_k/a_{k+1} - \zeta_{k+1}) > 0 gives convergence, and if 1/ζk\sum 1/\zeta_k diverges while that expression is eventually 0\le 0 the series diverges, and put

Rk  :=  (k+1)(akak+11)(kN),R_k \;:=\; (k+1)\left(\frac{a_k}{a_{k+1}} - 1\right) \qquad (k \in \mathbb{N}),

so that Kummer's expression for these weights is Kk=Rk1K_k = R_k - 1. Then:

  1. if lim infkRk>1\liminf_{k} R_k > 1 then ak\sum a_k converges;
  2. if lim supkRk<1\limsup_{k} R_k < 1 then ak\sum a_k diverges.

The weights are k+1k+1 rather than kk because ζ0\zeta_0 has to be positive and N\mathbb{N} contains 00; the classical statement, indexed from 11, is the same criterion read along the shift k=j+1k = j+1.

Nothing is claimed when lim infkRk1lim supkRk\liminf_k R_k \le 1 \le \limsup_k R_k. The Gauss test proved next is exactly the tool for the borderline case Rk1R_k \to 1, where Raabe's test is silent.

Facts & Assumptions

Given: A sequence (ak)(a_k) of reals with ak>0a_k > 0 for every kk; the weights ζk=ι(k+1)\zeta_k = \iota(k+1); and Rk=(k+1)(ak/ak+11)R_k = (k+1)(a_k/a_{k+1} - 1) (Limit superior and limit inferior of a real sequence as infnsupknxk\inf_n \sup_{k \ge n} x_k and supninfknxk\sup_n \inf_{k \ge n} x_k in R\overline{\mathbb{R}}, Canonical naturals are positive and strictly increasing).

[L3]

k11/kp\sum_{k \ge 1} 1/k^{p} converges if and only if p>1p > 1; at p=1p = 1 it therefore diverges (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1). Moreover k1=ι(k)k^{1} = \iota(k), the rational power at exponent 11 being the element itself (Rational powers ara^r of a positive base, Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a, Integer powers ama^m).

[L4]

The canonical naturals ι(k+1)\iota(k+1) are positive, and ι(k+2)=ι(k+1)+1\iota(k+2) = \iota(k+1) + 1 (Canonical naturals are positive and strictly increasing).

[L5]

The series k1xk\sum_{k \ge 1} x_k from the starting index 11 is by definition the series of the sequence jxj+1j \mapsto x_{j+1} (Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · direct
1.1

The weights ζk=ι(k+1)\zeta_k = \iota(k+1) are positive for every kNk \in \mathbb{N}, and the terms aka_k are positive, so Kummer's test applies with these data.

givenL4L2
1.2

Suppose lim infkRk>1\liminf_k R_k > 1. The real 11 is not an upper bound of the set of tail infima of (Rk)(R_k), so there is NN with iN>1i_N > 1, and iNi_N is real because iNRNi_N \le R_N.

givenL1choose
1.3

Suppose instead lim supkRk<1\limsup_k R_k < 1. The real 11 is not a lower bound of the set of tail suprema of (Rk)(R_k), so there is NN with sN<1s_N < 1, and then RksN<1R_k \le s_N < 1 for every kNk \ge N.

givenL1choose
2.1

Kummer's expression for these weights is Kk=(k+1)akak+1(k+2)=(k+1)(akak+11)1=Rk1K_k = (k+1)\dfrac{a_k}{a_{k+1}} - (k+2) = (k+1)\left(\dfrac{a_k}{a_{k+1}} - 1\right) - 1 = R_k - 1.

step 1.1L4algebra
2.2

The weight series is k1/ζk=k1/ι(k+1)\sum_k 1/\zeta_k = \sum_k 1/\iota(k+1), which is precisely the series k11/k\sum_{k \ge 1} 1/k from the starting index 11, and that is the case p=1p = 1 of the pp-series, hence divergent.

step 1.1L3L5
3.1

Put c:=iN1>0c := i_N - 1 > 0. For every kNk \ge N we have RkiNR_k \ge i_N, hence Kk=Rk1iN1=cK_k = R_k - 1 \ge i_N - 1 = c.

step 1.2step 2.1L1algebra
3.2

Hence Kk=Rk1<0K_k = R_k - 1 < 0, in particular Kk0K_k \le 0, for every kNk \ge N.

step 1.3step 2.1algebra
4.1

So cc is a lower bound of {Kk:kN}\{K_k : k \ge N\}, whence lim infkKkc>0\liminf_k K_k \ge c > 0, and Kummer's convergence criterion gives convergence of ak\sum a_k, which is claim 1.

step 3.1step 1.1L1L2
5.1

Together with the divergence of the weight series, Kummer's divergence criterion gives divergence of ak\sum a_k, which is claim 2.

step 3.2step 2.2step 1.1L2

Remarks

  • Raabe's test is a genuine strengthening of the ratio test. Whenever the ratios ak+1/aka_{k+1}/a_k converge to 11 the ratio test is silent, while RkR_k may still be bounded away from 11 on either side; the companion page carries a series with ratio limit exactly 11 that Raabe decides. The reason is visible in the weights: the divergent comparison series behind the test has moved from 1\sum 1 to the harmonic series, which diverges far more slowly.

  • The threshold is 11 and not 00, and step 2.1 says why. Kummer's criterion is a statement about Kk=Rk1K_k = R_k - 1; the shift by 11 between the two expressions is the whole difference between the two thresholds, and it comes from ζk+1ζk=1\zeta_{k+1} - \zeta_k = 1 for these weights.

Depends on

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