TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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.

If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value

Statement

Let $(x_{k})$ be a sequence of reals that converges (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), and let $(σ_{n})$ be its sequence of Cesaro means (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability). Then $(σ_{n})$ converges as well, and

$\lim_{n} σ_{n} = \lim_{k} x_{k} .$

Both limits are asserted to exist: the right-hand one by hypothesis, the left-hand one as part of the conclusion. Equivalently: a convergent sequence is $(C, 1)$ -summable, to its own limit. The notation is licensed by uniqueness of limits of real sequences (A sequence has at most one limit).

The converse is false (FALSE: if the Cesaro means of a sequence converge then the sequence converges).

Facts & Assumptions

Given: A sequence $(x_{k})$ of reals converging to $L : = \lim_{k} x_{k}$ , and its Cesaro means $σ_{n} = (n + 1)^{- 1} \sum_{k = 0}^{n} x_{k}$ .

[L1]

The Cesaro means and $(C, 1)$ -summability (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability); $(n + 1) \cdot 1_{R} > 0$ for every $n \in N$ (Canonical naturals are positive and strictly increasing).

[L2]

Finite sums (Finite sums and finite products, by recursion) and their laws: additivity, scaling with $\sum_{k < d} λ = d λ$ , splitting $\sum_{k < n} = \sum_{k < m} + \sum_{k = m}^{n - 1}$ for $m \leq n$ , and monotonicity of $\sum$ in its terms (Laws of finite sums and finite products).

[L3]

Triangle inequality for finite sums: $∣ \sum_{k < n} a_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣$ (Triangle inequality for finite sums).

[L4]

Convergence: for every rational $ε > 0$ there is $K$ with $∣ x_{k} - L ∣ < ε$ for all $k \geq K$ , and equivalently for every real $ε > 0$ , since below every positive real lies a positive rational (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, The rationals embed densely in the reals).

[L5]

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

[L6]

Order arithmetic: $a > 0$ gives $a^{- 1} > 0$ and $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); for $c > 0$ , $a \leq b$ gives $a c \leq b c$ (Sign rules for products and monotonicity of multiplication); adding a constant preserves the order and inequalities add (Order is preserved by adding a constant and by adding inequalities); the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field); $∣ u ∣ \geq 0$ (Basic properties of the absolute value); and $d \cdot 1_{R} \leq e \cdot 1_{R}$ whenever $d \leq e$ in $N$ (Canonical naturals are positive and strictly increasing). In each clause above, Sign rules for products and monotonicity of multiplication and Order is preserved by adding a constant and by adding inequalities state the STRICT forms and only those; the nonstrict forms used below are those together with the equality cases, which trichotomy settles, the order being total (Ordered field).

[L7]

The order on $N$ is total, so two indices have a larger one ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1L4L6choose

Let $ε > 0$ be an arbitrary real; choose $K \in N$ with $∣ x_{k} - L ∣ < ε / 2$ for every $k \geq K$ .

1.2L2L6

Put $C : = \sum_{k < K} ∣ x_{k} - L ∣$ , a real with $C \geq 0$ because every summand is $\geq 0$ .

2.1step 1.1step 1.2L1L2L3L6

For every $n \geq K$ : since $\sum_{k = 0}^{n} L = (n + 1) L$ , one has $σ_{n} - L = (n + 1)^{- 1} \sum_{k = 0}^{n} (x_{k} - L)$ , hence $∣ σ_{n} - L ∣ \leq (n + 1)^{- 1} \sum_{k = 0}^{n} ∣ x_{k} - L ∣ = (n + 1)^{- 1} (C + \sum_{k = K}^{n} ∣ x_{k} - L ∣) \leq (n + 1)^{- 1} (C + (n + 1 - K) (ε / 2)) \leq C (n + 1)^{- 1} + ε / 2$ , the last two steps using $∣ x_{k} - L ∣ < ε / 2$ for $K \leq k \leq n$ and $n + 1 - K \leq n + 1$ .

2.2step 1.2L5L6choose

Since $ε (2 (C + 1))^{- 1} > 0$ , choose a natural $m \geq 1$ with $1 / m < ε (2 (C + 1))^{- 1}$ ; then for every $n \geq m$ one has $n + 1 > m$ , so $C (n + 1)^{- 1} \leq C / m \leq (C + 1) / m < ε / 2$ .

3.1step 2.1step 2.2L6L7

Let $N$ be the larger of $K$ and $m$ ; for every $n \geq N$ both estimates apply and $∣ σ_{n} - L ∣ < ε / 2 + ε / 2 = ε$ .

4.1step 3.1L1L4∎

As $ε > 0$ was arbitrary, $(σ_{n})$ converges to $L$ , so $\lim_{n} σ_{n}$ exists and equals $L = \lim_{k} x_{k}$ .

Remarks

The shape of the argument is the shape of every regularity proof. The terms split into a fixed head, whose contribution is a constant divided by $n + 1$ and therefore eventually negligible, and a tail, all of whose terms are already within $ε / 2$ of $L$ and whose weights sum to at most $1$ . A summability matrix with only finitely many nonzero entries per row is regular iff each column tends to $0$ , the row sums tend to $1$ , and the row absolute sums are uniformly bounded is exactly this argument carried out for an arbitrary weighting, and The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem recovers the theorem above from it.
The bound $C + 1$ rather than $C$ in the second estimate is there only so that the divisor is positive when $C = 0$ , which happens whenever the first $K$ terms already equal $L$ .
Nothing here needs boundedness of $(x_{k})$ as a separate hypothesis: it follows from convergence, and in any case only the fixed head $x_{0}, \dots, x_{K - 1}$ is estimated crudely, and it is finite.

If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If xk→Lxk​→L then σn→Lσn​→L: convergence implies (C,1)(C,1)-summability to the same value

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If $x_{k} \to L$ then $σ_{n} \to L$ : convergence implies $(C, 1)$ -summability to the same value