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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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A series with ratio limit exactly 11 that Raabe decides

Example

Take ak:=1/ι(k+1)2a_k := 1/\iota(k+1)^{2} for kNk \in \mathbb{N}, so that ak\sum a_k is k11/k2\sum_{k \ge 1} 1/k^{2} (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

This is the smallest honest illustration that Raabe's test decides series the ratio test cannot. The verdict agrees with For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1 at p=2p = 2, as it must.

Facts & Assumptions

Given: The sequence ak:=1/ι(k+1)2a_k := 1/\iota(k+1)^{2}, kNk \in \mathbb{N}; its ratios qk=ak+1/akq_k = a_{k+1}/a_k; and its Raabe expression Rk=(k+1)(ak/ak+11)R_k = (k+1)(a_k/a_{k+1} - 1) (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, Integer powers ama^m, Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals are positive, so every aka_k is positive; reciprocation on the positives is order reversing; and x2=xxx^{2} = x\cdot x (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Integer powers ama^m, Monotonicity of xxnx \mapsto x^n and of nann \mapsto a^n).

[L2]

For every real ε>0\varepsilon > 0 there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, so 1/ι(k+1)01/\iota(k+1) \to 0 (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, Limits and Cauchy sequences of reals).

[L3]

Algebra of limits: sums, products and quotients of convergent sequences converge, the quotient requiring a nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients).

[L4]

The ratio test: its convergence half needs lim supkqk<1\limsup_k q_k < 1 and its divergence half needs lim infkqk>1\liminf_k q_k > 1 (Ratio test: lim supak+1/ak<1\limsup |a_{k+1}/a_k| < 1 gives absolute convergence and hence convergence, and lim infak+1/ak>1\liminf |a_{k+1}/a_k| > 1 gives divergence).

[L7]

k11/kp\sum_{k \ge 1} 1/k^{p} converges if and only if p>1p > 1 (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1).

Verification

technique · direct
1.1

Every ak=1/ι(k+1)2a_k = 1/\iota(k+1)^{2} is positive, so the ratios and the Raabe expression are defined.

givenL1
2.1

The ratios are qk=1/ι(k+2)21/ι(k+1)2=ι(k+1)2ι(k+2)2=(11ι(k+2))2q_k = \dfrac{1/\iota(k+2)^{2}}{1/\iota(k+1)^{2}} = \dfrac{\iota(k+1)^{2}}{\iota(k+2)^{2}} = \left(1 - \dfrac{1}{\iota(k+2)}\right)^{2}.

step 1.1L1algebra
2.2

The Raabe expression is Rk=(k+1)(ι(k+2)2ι(k+1)21)=ι(k+2)2ι(k+1)2ι(k+1)=2ι(k)+3ι(k)+1=2+1ι(k+1)R_k = (k+1)\left(\dfrac{\iota(k+2)^{2}}{\iota(k+1)^{2}} - 1\right) = \dfrac{\iota(k+2)^{2} - \iota(k+1)^{2}}{\iota(k+1)} = \dfrac{2\iota(k)+3}{\iota(k)+1} = 2 + \dfrac{1}{\iota(k+1)}.

step 1.1L1algebra
3.1

Since 1/ι(k+2)01/\iota(k+2) \to 0, the product rule gives qk(10)2=1q_k \to (1-0)^{2} = 1.

step 2.1L2L3
3.2

From step 2.2, Rk>2R_k > 2 for every kNk \in \mathbb{N}, the added term 1/ι(k+1)1/\iota(k+1) being positive.

step 2.2L1
4.1

The convergence half of the ratio test does not apply: if lim supkqk<1\limsup_k q_k < 1, then with tt real and lim supkqk<t<1\limsup_k q_k < t < 1 some tail supremum would be below tt, putting qkt<1q_k \le t < 1 for all large kk and contradicting qk1q_k \to 1.

step 3.1L4L6
4.2

The divergence half does not apply either: if lim infkqk>1\liminf_k q_k > 1, some tail infimum would exceed 11, putting qkc>1q_k \ge c > 1 for all large kk and again contradicting qk1q_k \to 1.

step 3.1L4L6
4.3

On the other hand 22 is a lower bound of {Rk:k0}\{R_k : k \ge 0\}, so the tail infimum i02i_0 \ge 2 and lim infkRk2>1\liminf_k R_k \ge 2 > 1.

step 3.2L6
5.1

Raabe's test therefore gives convergence of ak\sum a_k, that is of k11/k2\sum_{k \ge 1} 1/k^{2}, in agreement with the case p=2p = 2 of the pp-series theorem.

step 4.3step 1.1L5L7

Remarks

  • The Raabe expression here is exact, not asymptotic. Step 2.2 computes Rk=2+1/(k+1)R_k = 2 + 1/(k+1) on the nose, so no limit is needed to apply the test: a single inequality Rk>2R_k > 2 at every index already forces lim infkRk2\liminf_k R_k \ge 2. That is why this witness is the cleanest available one.

  • Why the ratio test must fail here. The ratios of any pp-series tend to 11 whatever pp is, so a criterion reading only lim sup\limsup and lim inf\liminf of the ratios cannot separate the convergent pp-series from the divergent ones. Raabe reads the rate at which the ratios approach 11, which is exactly the missing information, and that rate is 2/k2/k up to smaller terms when p=2p = 2.

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