ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The period-three pattern $1, 1, - 2$ has partial sums in ${0, 1, 2}$ , so $\sum a_{k} / (k + 1)$ converges by Dirichlet's test although the alternating series test does not apply

Example

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

$a_{k} : = {\begin{cases} 1 & if g_{k} \in {0, 1}, \\ - 2 & if g_{k} = 2, \end{cases} b_{k} : = \frac{1}{ι (k + 1)} .$

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

$\sum_{k} \frac{a_{k}}{k + 1}$

converges. It converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index), since $∣ a_{k} ∣ \geq 1$ for every $k$ and $\sum 1 / ι (k + 1)$ is the harmonic series.

The alternating series test does not reach this example. The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ is a statement about $\sum ε_{k} b_{k}$ for the alternating sequence $(ε_{k})$ , whose values strictly alternate in sign; here $a_{0} = a_{1} = 1$ , so $(a_{k})$ 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 $(g_{k})$ with values in ${0, 1, 2}$ defined by the displayed recursion, the terms $a_{k} \in {1, - 2}$ read off from it, $b_{k} = 1 / ι (k + 1)$ , and the partial sums $A_{n} = \sum_{k < n} a_{k}$ .

[L1]

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

[L2]

Partial sums satisfy $A_{0} = 0$ and $A_{n + 1} = A_{n} + a_{n}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L3]

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

[L4]

Dirichlet's test: bounded partial sums of $\sum a_{k}$ and a nonincreasing $(b_{k})$ with $b_{k} \to 0$ give convergence of $\sum a_{k} b_{k}$ (Dirichlet's test: if the partial sums of $\sum a_{k}$ are bounded and $(b_{k})$ is nonincreasing with $b_{k} \to 0$ , then $\sum a_{k} b_{k}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set, Limits and Cauchy sequences of reals).

[L5]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ , with $ι (k)^{1} = ι (k)$ ; and $\sum_{k \geq 1} x_{k}$ is the series of $j \mapsto x_{j + 1}$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Integer powers $a^{m}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

Direct comparison, in its divergence form: if $0 \leq x_{k} \leq y_{k}$ from some index on and $\sum x_{k}$ diverges then $\sum y_{k}$ diverges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L7]

Absolute value: $∣ x ∣ \geq 0$ and $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ (Basic properties of the absolute value).

[L8]

Absolute and conditional convergence (Absolutely convergent and conditionally convergent series, and the general starting index).

[L9]

The alternating series test is stated for $\sum ε_{k} b_{k}$ with $(ε_{k})$ the alternating sequence, which satisfies $ε_{0} = 1$ and $ε_{k + 1} = - ε_{k}$ (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

Verification

technique · direct

1.1

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

givenL1

1.2

Every $b_{k} = 1 / ι (k + 1)$ is positive, $(b_{k})$ is nonincreasing since $0 < ι (k + 1) < ι (k + 2)$ , and $b_{k} \to 0$ : given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε$ gives $b_{k} \leq 1 / ι (n) < ε$ for every $k \geq n$ .

givenL3

2.1

An induction gives $A_{n} = ι (g_{n})$ for every $n$ : at $n = 0$ both sides are $0$ ; and if $A_{n} = ι (g_{n})$ then, when $g_{n} \in {0, 1}$ we have $a_{n} = 1$ and $g_{n + 1} = g_{n} + 1$ , so $A_{n + 1} = ι (g_{n}) + 1 = ι (g_{n + 1})$ , while when $g_{n} = 2$ we have $a_{n} = - 2$ and $g_{n + 1} = 0$ , so $A_{n + 1} = ι (2) - 2 = 0 = ι (g_{n + 1})$ .

step 1.1L1L2

2.2

For every $k$ , $∣ a_{k} b_{k} ∣ = ∣ a_{k} ∣ b_{k} \geq b_{k} = 1 / ι (k + 1) \geq 0$ , since $∣ a_{k} ∣$ is $1$ or $2$ .

step 1.1step 1.2L7

3.1

Hence $A_{n} \in {0, 1, 2}$ for every $n$ and $∣ A_{n} ∣ \leq 2$ : the range of the partial sums is bounded.

step 2.1L7

3.2

The series $\sum_{k} 1 / ι (k + 1)$ is the $p$ -series $\sum_{k \geq 1} 1 / k$ at $p = 1$ , which diverges; so by comparison $\sum_{k} ∣ a_{k} b_{k} ∣$ diverges.

step 2.2L5L6

4.1

By Dirichlet's test, $\sum_{k} a_{k} b_{k}$ converges.

step 3.1step 1.2L4

5.1

Therefore $\sum_{k} a_{k} b_{k}$ 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 $(ε_{k})$ , for which $ε_{0} = 1$ and $ε_{1} = - 1$ , whereas here $a_{0} = a_{1} = 1$ .

step 1.1L9∎

Remarks

Dirichlet's test needs only boundedness of the partial sums. Here they cycle through $0, 1, 2$ and never converge, so $\sum a_{k}$ itself diverges; the test nevertheless applies, and that is precisely what distinguishes it from Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges, which requires $\sum a_{k}$ to converge.
Why the pattern is $1, 1, - 2$ and not $1, - 1$ . The blocks must have mean zero for the partial sums to stay bounded, and they must not alternate, or the example would be covered by the alternating series test. The smallest integer pattern with both properties has period three.
The parity object is not used, and could not be. The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ supplies the period-two structure of this library; a period-three pattern needs its own recursion, which is what the state set ${0, 1, 2}$ provides. No claim is made that the two constructions are instances of a common one.

Depends on

Used by

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Dependency tree · two levels

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Sources

Dirichlet's test (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)

The period-three pattern 1,1,−21,1,−2 has partial sums in {0,1,2}{0,1,2}, so ∑ak/(k+1)∑ak​/(k+1) converges by Dirichlet's test although the alternating series test does not apply