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CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem

Statement

Define cn,k:=(n+1)1c_{n,k} := (n+1)^{-1} for knk \le n and cn,k:=0c_{n,k} := 0 for k>nk > n. Then:

  1. cc is a summability matrix (A summability (Toeplitz) matrix, the transformed sequence yn=kcn,kxky_n = \sum_k c_{n,k} x_k, and regularity), with nn an admissible bound for row nn;
  2. the transform of a sequence (xk)(x_k) by cc is exactly its sequence of Cesaro means (The Cesaro means σn=(x0++xn)/(n+1)\sigma_n = (x_0 + \dots + x_n)/(n+1) and (C,1)(C,1)-summability), yn=σny_n = \sigma_n;
  3. cc satisfies the three conditions of A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to 00, the row sums tend to 11, and the row absolute sums are uniformly bounded, so cc is regular.

Consequently every convergent sequence has σnlimkxk\sigma_n \to \lim_k x_k, which is a second proof of If xkLx_k \to L then σnL\sigma_n \to L: convergence implies (C,1)(C,1)-summability to the same value, obtained from the general characterisation rather than from a direct estimate.

Facts & Assumptions

Given: The matrix cc with cn,k=(n+1)1c_{n,k} = (n+1)^{-1} for knk \le n and cn,k=0c_{n,k} = 0 for k>nk > n.

[L2]

The Cesaro means σn=(n+1)1k=0nxk\sigma_n = (n+1)^{-1}\sum_{k=0}^{n} x_k (The Cesaro means σn=(x0++xn)/(n+1)\sigma_n = (x_0 + \dots + x_n)/(n+1) and (C,1)(C,1)-summability).

[L3]

Silverman-Toeplitz: a summability matrix is regular exactly when every column tends to 00, the row sums tend to 11, and the row absolute sums are uniformly bounded (A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to 00, the row sums tend to 11, and the row absolute sums are uniformly bounded).

[L4]

Finite sums and their laws, in particular k<dλ=dλ\sum_{k<d}\lambda = d\lambda (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Convergence, and the fact that a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

Reciprocal Archimedean property: for every real ε>0\varepsilon > 0 there is a natural m1m \ge 1 with 1/m<ε1/m < \varepsilon (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon).

[L7]

Order arithmetic: (n+1)1R>0(n+1)\cdot 1_{\mathbb{R}} > 0 for every nNn \in \mathbb{N} (Canonical naturals are positive and strictly increasing); a positive element is invertible with positive inverse, and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order); u=u|u| = u for u0u \ge 0 (Basic properties of the absolute value); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct
1.1

Row nn of cc vanishes at every k>nk > n, so nn is an admissible bound for row nn and cc is a summability matrix; its transform is yn=k=0n(n+1)1xk=(n+1)1k=0nxk=σny_n = \sum_{k=0}^{n}(n+1)^{-1}x_k = (n+1)^{-1}\sum_{k=0}^{n}x_k = \sigma_n.

L1L2L4
1.2

For every nn the canonical natural (n+1)1R(n+1)\cdot 1_{\mathbb{R}} is positive, hence invertible with (n+1)1>0(n+1)^{-1} > 0; so cn,k=cn,k|c_{n,k}| = c_{n,k} for all n,kn,k.

L7
2.1

Columns are null. Fix kk and let ε>0\varepsilon > 0; choose m1m \ge 1 with 1/m<ε1/m < \varepsilon. For nmn \ge m one has n+1>mn + 1 > m, so cn,k0(n+1)1<1/m<ε|c_{n,k} - 0| \le (n+1)^{-1} < 1/m < \varepsilon, the case n<kn < k giving cn,k=0c_{n,k} = 0 outright. Hence limncn,k=0\lim_n c_{n,k} = 0.

step 1.2L5L6L7
2.2

Row sums tend to 11. For every nn, kcn,k=k=0n(n+1)1=(n+1)(n+1)1=1\sum_k c_{n,k} = \sum_{k=0}^{n}(n+1)^{-1} = (n+1)(n+1)^{-1} = 1, a constant sequence, which converges to 11.

step 1.1step 1.2L1L4L5
2.3

Row absolute sums are uniformly bounded. For every nn, kcn,k=kcn,k=11\sum_k |c_{n,k}| = \sum_k c_{n,k} = 1 \le 1.

step 1.1step 1.2L1L4
3.1

All three conditions hold, so cc is regular.

step 2.1step 2.2step 2.3L3
4.1

Therefore, for every convergent sequence (xk)(x_k), the transform (σn)(\sigma_n) converges with limnσn=limkxk\lim_n \sigma_n = \lim_k x_k.

step 1.1step 3.1L1

Remarks

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