False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: if the Cesaro means of a sequence converge then the sequence converges

Statement

False claim: if the Cesaro means $(\sigma_n)$ of a sequence $(x_k)$ of reals converge (The Cesaro means $\sigma_n = (x_0 + \dots + x_n)/(n+1)$ and $(C,1)$ -summability), then $(x_k)$ converges.

The implication in the opposite direction is true and is If $x_k \to L$ then $\sigma_n \to L$ : convergence implies $(C,1)$ -summability to the same value. The claim above asserts its converse, and it is refuted below by the alternating sequence $s_k = (-1)^k$ , whose Cesaro means converge to $0$ while the sequence itself does not converge at all.

That is the whole reason Cesaro summability is worth defining: it is a strictly larger notion than convergence, consistent with it where both apply.

Facts & Assumptions

Given: The alternating sequence $(s_k)$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ , the unique sequence of reals with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ , together with the index maps $e$ and $o$ of that lemma; and its partial sums $S_n := \sum_{k<n} s_k$ (Finite sums and finite products, by recursion), and its Cesaro means $\sigma_n = (n+1)^{-1}S_{n+1}$ (The Cesaro means $\sigma_n = (x_0 + \dots + x_n)/(n+1)$ and $(C,1)$ -summability).

[L1]

The alternating sequence: $e$ and $o$ are the unique maps with $e_0 = 0$ , $e_{\sigma(j)} = \sigma(\sigma(e_j))$ , $o_0 = \sigma(0)$ , $o_{\sigma(j)} = \sigma(\sigma(o_j))$ ; both are strictly increasing; $\mathbb{N}$ is the disjoint union of their ranges; $(s_k)$ is the unique sequence with $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ ; $|s_k| = 1$ , $s_{e_j} = 1$ and $s_{o_j} = -1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ).

[L2]

Finite sums: $\sum_{k<0} s_k = 0$ and $\sum_{k<n+1} s_k = \sum_{k<n} s_k + s_n$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L3]

Induction principle (The principle of mathematical induction).

[L4]

Cesaro means: $\sigma_n = (n+1)^{-1}\sum_{k=0}^{n}x_k$ , and $\sum_{k=0}^{n} = \sum_{k<n+1}$ (The Cesaro means $\sigma_n = (x_0 + \dots + x_n)/(n+1)$ and $(C,1)$ -summability, Finite sums and finite products, by recursion).

[L5]

The alternating sequence does not converge: it is bounded and divergent, which is the refutation of FALSE: every bounded sequence converges, carried out there for the very same sequence, the one determined by $s_0 = 1$ and $s_{\sigma(k)} = -s_k$ , which The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ shows is unique.

[L6]

Reciprocal Archimedean property (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ); convergence of a real sequence (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

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

Refutation

technique · direct

1.1

By induction on $j$ , $o_j = \sigma(e_j)$ and $e_{\sigma(j)} = \sigma(o_j)$ : at $j = 0$ one has $o_0 = \sigma(0) = \sigma(e_0)$ , and $e_{\sigma(j)} = \sigma(\sigma(e_j)) = \sigma(o_j)$ follows from the first identity at $j$ , while $o_{\sigma(j)} = \sigma(\sigma(o_j)) = \sigma(e_{\sigma(j)})$ carries the first identity to $\sigma(j)$ .

L1L3

1.2

The partial sums satisfy $S_0 = 0$ and $S_{\sigma(n)} = S_n + s_n$ , and $\sigma_n = (n+1)^{-1}S_{\sigma(n)}$ .

L2L4L7

1.3

$(s_k)$ does not converge.

2.1

By induction on $j$ : $S_{e_j} = 0$ and $S_{o_j} = 1$ . At $j = 0$ : $S_{e_0} = S_0 = 0$ and $S_{o_0} = S_{\sigma(0)} = S_0 + s_0 = 1$ . For the step, $S_{e_{\sigma(j)}} = S_{\sigma(o_j)} = S_{o_j} + s_{o_j} = 1 + (-1) = 0$ and $S_{o_{\sigma(j)}} = S_{\sigma(e_{\sigma(j)})} = S_{e_{\sigma(j)}} + s_{e_{\sigma(j)}} = 0 + 1 = 1$ .

step 1.1step 1.2L1L3

3.1

Every natural number is $e_j$ for exactly one $j$ or $o_j$ for exactly one $j$ , so $S_m \in \{0,1\}$ for every $m$ ; in particular $0 \le S_{\sigma(n)} \le 1$ for every $n$ .

step 1.1step 2.1L1

4.1

Hence $0 \le \sigma_n = (n+1)^{-1}S_{\sigma(n)} \le (n+1)^{-1}$ and so $|\sigma_n| \le (n+1)^{-1}$ for every $n$ .

step 1.2step 3.1L7

5.1

Given a real $\varepsilon > 0$ , choose $m \ge 1$ with $1/m < \varepsilon$ ; for every $n \ge m$ one has $n+1 > m$ and therefore $|\sigma_n - 0| \le (n+1)^{-1} < 1/m < \varepsilon$ . So $(\sigma_n)$ converges to $0$ , that is, $(s_k)$ is $(C,1)$ -summable to $0$ .

step 4.1L4L6L7

6.1

So $(s_k)$ has convergent Cesaro means and does not converge, and the claim is false.

step 1.3step 5.1∎

Remarks

What averaging destroys. The Cesaro mean of the first $n+1$ terms of an alternating sequence is either $0$ or $1/(n+1)$ , because the terms cancel in pairs and at most one is left over. The oscillation is real and is not damped by any tail condition; it is simply invisible to the average. So the transform loses information, and no regular summability method can be expected to recover a limit that does not exist (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem).
The worked computation of the means, with the values displayed, is The Cesaro means of $(-1)^k$ converge to $0$ although the sequence diverges ↗.
A correct converse needs an extra hypothesis. The classical one is Tauberian: if the Cesaro means converge and in addition $k(x_k - x_{k-1})$ is bounded, then $(x_k)$ converges. No such theorem is proved in this library, and none may be cited from it; the statement is mentioned only to say what the repaired claim would look like.
The failure is not caused by unboundedness. The witness is bounded, with $|s_k| = 1$ at every index. It is the same sequence that refutes the claim that bounded sequences converge (FALSE: every bounded sequence converges), and for the same underlying reason: boundedness forbids escaping, not oscillating.

Depends on

Used by

The Cesaro means of (-1)ᵏ converge to 0 although the sequence diverges Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 16 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

Grandi's series (Wikipedia) (standard reference, not scraped)
Cesàro summation (Wikipedia) (standard reference, not scraped)
Divergent series (Wikipedia) (standard reference, not scraped)
G. H. Hardy, Divergent Series, Ch. 1 (standard reference, not scraped)