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ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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The period-three pattern 1,1,21, 1, -2 has partial sums in {0,1,2}\{0,1,2\}, so ak/(k+1)\sum a_k/(k+1) converges by Dirichlet's test although the alternating series test does not apply

Example

Let (gk)(g_k) be the sequence of naturals with values in {0,1,2}\{0,1,2\} defined by the recursion g0=0g_0 = 0 and gk+1=gk+1g_{k+1} = g_k + 1 for gk{0,1}g_k \in \{0,1\}, gk+1=0g_{k+1} = 0 for gk=2g_k = 2 (The recursion theorem), and put

ak:={1if gk{0,1},2if gk=2,bk:=1ι(k+1).a_k := \begin{cases} 1 & \text{if } g_k \in \{0,1\}, \\ -2 & \text{if } g_k = 2, \end{cases} \qquad b_k := \frac{1}{\iota(k+1)} .

So (ak)(a_k) is the repeating pattern 1,1,2,1,1,2,1, 1, -2, 1, 1, -2, \dots Its partial sums An=k<nakA_n = \sum_{k<n} a_k take only the values 0,1,20, 1, 2, hence are bounded (Lower bound, bounded below, bounded set), while (bk)(b_k) is nonincreasing with bk0b_k \to 0. By Dirichlet's test (Dirichlet's test: if the partial sums of ak\sum a_k are bounded and (bk)(b_k) is nonincreasing with bk0b_k \to 0, then akbk\sum a_k b_k converges) the series

kakk+1\sum_k \frac{a_k}{k+1}

converges. It converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index), since ak1|a_k| \ge 1 for every kk and 1/ι(k+1)\sum 1/\iota(k+1) is the harmonic series.

The alternating series test does not reach this example. The alternating series test: if (bk)(b_k) is nonincreasing with bk0b_k \to 0 then k(1)kbk\sum_{k} (-1)^{k} b_k converges, the sum lies between any two consecutive partial sums, and the error after nn terms is at most bnb_n is a statement about εkbk\sum \varepsilon_k b_k for the alternating sequence (εk)(\varepsilon_k), whose values strictly alternate in sign; here a0=a1=1a_0 = a_1 = 1, so (ak)(a_k) is not that sequence, nor any constant multiple of it, and no reading of the test applies. This is the item on the page showing that Dirichlet's test is strictly stronger than the Leibniz criterion, and an alternating witness would not show it.

Facts & Assumptions

Given: The sequence (gk)(g_k) with values in {0,1,2}\{0,1,2\} defined by the displayed recursion, the terms ak{1,2}a_k \in \{1,-2\} read off from it, bk=1/ι(k+1)b_k = 1/\iota(k+1), and the partial sums An=k<nakA_n = \sum_{k<n} a_k.

[L1]

The recursion theorem and the principle of induction (The recursion theorem, The principle of mathematical induction).

[L3]

The canonical naturals are positive for n1n \ge 1 and strictly increasing; reciprocation reverses the order on the positives; and for every real ε>0\varepsilon > 0 there is n1n \ge 1 with 1/ι(n)<ε1/\iota(n) < \varepsilon (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon).

[L5]

k11/kp\sum_{k\ge1}1/k^{p} converges if and only if p>1p > 1, with ι(k)1=ι(k)\iota(k)^{1} = \iota(k); and k1xk\sum_{k\ge1}x_k is the series of jxj+1j \mapsto x_{j+1} (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1, 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, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

Direct comparison, in its divergence form: if 0xkyk0 \le x_k \le y_k from some index on and xk\sum x_k diverges then yk\sum y_k diverges (If 0akbk0 \le a_k \le b_k eventually, convergence of bk\sum b_k gives convergence of ak\sum a_k, and divergence of ak\sum a_k gives divergence of bk\sum b_k).

[L7]

Absolute value: x0|x| \ge 0 and xy=xy|xy| = |x|\,|y| (Basic properties of the absolute value).

Verification

technique · direct
1.1

The recursion defines (gk)(g_k) as a function N{0,1,2}\mathbb{N} \to \{0,1,2\}, the transition being a total function of the set {0,1,2}\{0,1,2\} to itself; hence (ak)(a_k) is a well-defined sequence of reals with values in {1,2}\{1, -2\}.

givenL1
1.2

Every bk=1/ι(k+1)b_k = 1/\iota(k+1) is positive, (bk)(b_k) is nonincreasing since 0<ι(k+1)<ι(k+2)0 < \iota(k+1) < \iota(k+2), and bk0b_k \to 0: given a rational ε>0\varepsilon > 0, an n1n \ge 1 with 1/ι(n)<ε1/\iota(n) < \varepsilon gives bk1/ι(n)<εb_k \le 1/\iota(n) < \varepsilon for every knk \ge n.

givenL3
2.1

An induction gives An=ι(gn)A_n = \iota(g_n) for every nn: at n=0n = 0 both sides are 00; and if An=ι(gn)A_n = \iota(g_n) then, when gn{0,1}g_n \in \{0,1\} we have an=1a_n = 1 and gn+1=gn+1g_{n+1} = g_n + 1, so An+1=ι(gn)+1=ι(gn+1)A_{n+1} = \iota(g_n) + 1 = \iota(g_{n+1}), while when gn=2g_n = 2 we have an=2a_n = -2 and gn+1=0g_{n+1} = 0, so An+1=ι(2)2=0=ι(gn+1)A_{n+1} = \iota(2) - 2 = 0 = \iota(g_{n+1}).

step 1.1L1L2
2.2

For every kk, akbk=akbkbk=1/ι(k+1)0|a_k b_k| = |a_k| b_k \ge b_k = 1/\iota(k+1) \ge 0, since ak|a_k| is 11 or 22.

step 1.1step 1.2L7
3.1

Hence An{0,1,2}A_n \in \{0, 1, 2\} for every nn and An2|A_n| \le 2: the range of the partial sums is bounded.

step 2.1L7
3.2

The series k1/ι(k+1)\sum_k 1/\iota(k+1) is the pp-series k11/k\sum_{k\ge1}1/k at p=1p = 1, which diverges; so by comparison kakbk\sum_k |a_k b_k| diverges.

step 2.2L5L6
4.1

By Dirichlet's test, kakbk\sum_k a_k b_k converges.

step 3.1step 1.2L4
5.1

Therefore kakbk\sum_k a_k b_k converges conditionally: it converges by step 4.1 and does not converge absolutely by step 3.2.

step 4.1step 3.2L8
6.1

The alternating series test does not apply to this series: it is a statement about the alternating sequence (εk)(\varepsilon_k), for which ε0=1\varepsilon_0 = 1 and ε1=1\varepsilon_1 = -1, whereas here a0=a1=1a_0 = a_1 = 1.

step 1.1L9

Remarks

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