CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$1 - 2 + 3 - 4 + \dots$ is Abel summable to $1 / 4$ but is not Cesaro summable

Statement

The series $\sum_{n \geq 0} (- 1)^{n} ι (n + 1)$ is Abel summable to $1 / 4$ , but its Cesaro means do not converge.

Facts & Assumptions

Given: Coefficients $a_{n} = (- 1)^{n} ι (n + 1)$ .

[L1]

The geometric series for $1 / (1 + x)$ may be differentiated term by term for $∣ x ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Inside its radius a real power series may be differentiated term by term, and the differentiated series has the same radius).

[L2]

Cesaro means average the inclusive partial sums (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability, Abel summability by $\lim_{x ↑ 1} \sum a_{n} x^{n}$ and Cesaro summability by the Cesaro means of the partial sums).

Verification

technique · direct

1.1

Differentiating $\sum_{n \geq 0} (- 1)^{n} x^{n} = 1 / (1 + x)$ and combining with the original series gives $\sum_{n \geq 0} (- 1)^{n} ι (n + 1) x^{n} = 1 / (1 + x)^{2}$ for $0 \leq x < 1$ . Its limit as $x ↑ 1$ is $1 / 4$ .

L1algebra

1.2

The inclusive partial sums satisfy $S_{2 m} = ι (m + 1)$ and $S_{2 m + 1} = - ι (m + 1)$ . Hence $σ_{2 m} = ι (m + 1) / ι (2 m + 1) \to 1 / 2$ , while $σ_{2 m + 1} = 0$ .

L2algebra

2.1

Thus the Cesaro means have two distinct subsequential limits and do not converge, whereas step 1.1 proves Abel summability to $1 / 4$ .

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Abel summability, Encyclopedia of Mathematics (standard reference, not scraped)
Cesàro summation, Encyclopedia of Mathematics (standard reference, not scraped)
Cesàro summation methods, Encyclopedia of Mathematics (standard reference, not scraped)
S. Semmes, Rice Math 322 notes (standard reference, not scraped)

1−2+3−4+⋯1−2+3−4+⋯ is Abel summable to 1/41/4 but is not Cesaro summable