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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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If xk→L then σn→L: convergence implies (C,1)-summability to the same value

Statement

Let (xk) be a sequence of reals that converges (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), and let (σn) be its sequence of Cesaro means (The Cesaro means σn=(x0+⋯+xn)/(n+1) and (C,1)-summability). Then (σn) converges as well, and

lim⁡nσn  =  lim⁡kxk.

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 (xk) of reals converging to L:=lim⁡kxk, and its Cesaro means σn=(n+1)−1∑k=0nxk.

[L1]

The Cesaro means and (C,1)-summability (The Cesaro means σn=(x0+⋯+xn)/(n+1) and (C,1)-summability); (n+1)⋅1R>0 for every n∈N (Canonical naturals are positive and strictly increasing).

[L2]

Finite sums (Finite sums and finite products, by recursion) and their laws: additivity, scaling with ∑k<dλ=dλ, splitting ∑k<n=∑k<m+∑k=mn−1 for m≤n, and monotonicity of ∑ in its terms (Laws of finite sums and finite products).

[L3]

Triangle inequality for finite sums: ∣∑k<nak∣≤∑k<n∣ak∣ (Triangle inequality for finite sums).

[L4]

Convergence: for every rational ε>0 there is K with ∣xk−L∣<ε for all k≥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≥1 with 1/m<ε (For every ε>0 in a complete ordered field there is a natural n≥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≤b gives ac≤bc (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∣≥0 (Basic properties of the absolute value); and d⋅1R≤e⋅1R whenever d≤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 (≤ is a linear order on N).

Proof

technique · direct
1.1

Let ε>0 be an arbitrary real; choose K∈N with ∣xk−L∣<ε/2 for every k≥K.

L4L6choose
1.2

Put C:=∑k<K∣xk−L∣, a real with C≥0 because every summand is ≥0.

L2L6
2.1

For every n≥K: since ∑k=0nL=(n+1)L, one has σn−L=(n+1)−1∑k=0n(xk−L), hence ∣σn−L∣≤(n+1)−1∑k=0n∣xk−L∣=(n+1)−1(C+∑k=Kn∣xk−L∣)≤(n+1)−1(C+(n+1−K)(ε/2))≤C(n+1)−1+ε/2, the last two steps using ∣xk−L∣<ε/2 for K≤k≤n and n+1−K≤n+1.

step 1.1step 1.2L1L2L3L6
2.2

Since ε (2(C+1))−1>0, choose a natural m≥1 with 1/m<ε (2(C+1))−1; then for every n≥m one has n+1>m, so C(n+1)−1≤C/m≤(C+1)/m<ε/2.

step 1.2L5L6choose
3.1

Let N be the larger of K and m; for every n≥N both estimates apply and ∣σn−L∣<ε/2+ε/2=ε.

step 2.1step 2.2L6L7
4.1

As ε>0 was arbitrary, (σn) converges to L, so lim⁡nσn exists and equals L=lim⁡kxk.

step 3.1L1L4∎

Remarks

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