Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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+⋯1-2+3-4+\cdots is Abel summable to 1/41/4 but is not Cesaro summable

Statement

The series ∑n≥0(−1)nι(n+1)\sum_{n\ge0}(-1)^n\iota(n+1) is Abel summable to 1/41/4, but its Cesaro means do not converge.

Facts & Assumptions

Given: Coefficients an=(−1)nι(n+1)a_n=(-1)^n\iota(n+1).

[L1]

The geometric series for 1/(1+x)1/(1+x) may be differentiated term by term for ∣x∣<1|x|<1 (For ∣r∣<1|r| < 1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r), and for ∣r∣≥1|r| \ge 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 σn=(x0+⋯+xn)/(n+1)\sigma_n = (x_0 + \dots + x_n)/(n+1) and (C,1)(C,1)-summability, Abel summability by lim⁡x↑1∑anxn\lim_{x\uparrow1}\sum a_nx^n and Cesaro summability by the Cesaro means of the partial sums).

Verification

technique · direct
1.1

Differentiating ∑n≥0(−1)nxn=1/(1+x)\sum_{n\ge0}(-1)^nx^n=1/(1+x) and combining with the original series gives ∑n≥0(−1)nι(n+1)xn=1/(1+x)2\sum_{n\ge0}(-1)^n\iota(n+1)x^n=1/(1+x)^2 for 0≤x<10\le x<1. Its limit as x↑1x\uparrow1 is 1/41/4.

L1algebra
1.2

The inclusive partial sums satisfy S2m=ι(m+1)S_{2m}=\iota(m+1) and S2m+1=−ι(m+1)S_{2m+1}=-\iota(m+1). Hence σ2m=ι(m+1)/ι(2m+1)→1/2\sigma_{2m}=\iota(m+1)/\iota(2m+1)\to1/2, while σ2m+1=0\sigma_{2m+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 1/41/4.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 95 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources