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Raabe is Kummer with ζk=k+1: for positive terms, lim inf⁡ (k+1)(ak/ak+1−1)>1 gives convergence and lim sup⁡<1 gives divergence

Statement

Let (ak) be a sequence of reals with ak>0 for every k∈N. Write k+1 for the canonical natural ι(k+1)∈R, which is positive (Canonical naturals are positive and strictly increasing), take the weights ζk:=k+1 in Kummer: for positive terms ak and weights ζk>0, lim inf⁡(ζkak/ak+1−ζk+1)>0 gives convergence, and if ∑1/ζk diverges while that expression is eventually ≤0 the series diverges, and put

Rk  :=  (k+1)(akak+1−1)(k∈N),

so that Kummer's expression for these weights is Kk=Rk−1. Then:

  1. if lim inf⁡kRk>1 then ∑ak converges;
  2. if lim sup⁡kRk<1 then ∑ak diverges.

The weights are k+1 rather than k because ζ0 has to be positive and N contains 0; the classical statement, indexed from 1, is the same criterion read along the shift k=j+1.

Nothing is claimed when lim inf⁡kRk≤1≤lim sup⁡kRk. The Gauss test proved next is exactly the tool for the borderline case Rk→1, where Raabe's test is silent.

Facts & Assumptions

Given: A sequence (ak) of reals with ak>0 for every k; the weights ζk=ι(k+1); and Rk=(k+1)(ak/ak+1−1) (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾, Canonical naturals are positive and strictly increasing).

[L3]

∑k≥11/kp converges if and only if p>1; at p=1 it therefore diverges (For rational p>0, ∑1/kp converges iff p>1). Moreover k1=ι(k), the rational power at exponent 1 being the element itself (Rational powers ar of a positive base, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Integer powers am).

[L4]

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

[L5]

The series ∑k≥1xk from the starting index 1 is by definition the series of the sequence j↦xj+1 (Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · direct
1.1

The weights ζk=ι(k+1) are positive for every k∈N, and the terms ak are positive, so Kummer's test applies with these data.

givenL4L2
1.2

Suppose lim inf⁡kRk>1. The real 1 is not an upper bound of the set of tail infima of (Rk), so there is N with iN>1, and iN is real because iN≤RN.

givenL1choose
1.3

Suppose instead lim sup⁡kRk<1. The real 1 is not a lower bound of the set of tail suprema of (Rk), so there is N with sN<1, and then Rk≤sN<1 for every k≥N.

givenL1choose
2.1

Kummer's expression for these weights is Kk=(k+1)akak+1−(k+2)=(k+1)(akak+1−1)−1=Rk−1.

step 1.1L4algebra
2.2

The weight series is ∑k1/ζk=∑k1/ι(k+1), which is precisely the series ∑k≥11/k from the starting index 1, and that is the case p=1 of the p-series, hence divergent.

step 1.1L3L5
3.1

Put c:=iN−1>0. For every k≥N we have Rk≥iN, hence Kk=Rk−1≥iN−1=c.

step 1.2step 2.1L1algebra
3.2

Hence Kk=Rk−1<0, in particular Kk≤0, for every k≥N.

step 1.3step 2.1algebra
4.1

So c is a lower bound of {Kk:k≥N}, whence lim inf⁡kKk≥c>0, and Kummer's convergence criterion gives convergence of ∑ak, 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, 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/ak converge to 1 the ratio test is silent, while Rk may still be bounded away from 1 on either side; the companion page carries a series with ratio limit exactly 1 that Raabe decides. The reason is visible in the weights: the divergent comparison series behind the test has moved from ∑1 to the harmonic series, which diverges far more slowly.

  • The threshold is 1 and not 0, and step 2.1 says why. Kummer's criterion is a statement about Kk=Rk−1; the shift by 1 between the two expressions is the whole difference between the two thresholds, and it comes from ζk+1−ζk=1 for these weights.

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