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The Cesaro means of $(- 1)^{k}$ converge to $0$ although the sequence diverges

Example

Let $(s_{k})$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , 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 $s_{σ (k)} = - s_{k}$ , usually written $s_{k} = (- 1)^{k}$ . Its Cesaro means (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability) are

$σ_{n} = \frac{s_{0} + \dots + s_{n}}{n + 1} = {\begin{cases} \frac{1}{n + 1} & n even, \\ 0 & n odd, \end{cases}$

so the first few values are

$σ_{0} = 1, σ_{1} = 0, σ_{2} = \frac{1}{3}, σ_{3} = 0, σ_{4} = \frac{1}{5}, σ_{5} = 0, \dots$

and $\lim_{n} σ_{n} = 0$ , while $(s_{k})$ does not converge at all. So $(s_{k})$ is $(C, 1)$ -summable to $0$ and divergent: it is the standard witness that $(C, 1)$ -summability is strictly weaker than convergence, and the one used in FALSE: if the Cesaro means of a sequence converge then the sequence converges.

The value $0$ is the one an average ought to give, since the sequence spends half its indices at $1$ and half at $- 1$ ; the classical way to say this is that the series $1 - 1 + 1 - 1 + \dots$ has Cesaro sum $\frac{1}{2}$ , that being the Cesaro limit of its partial sums rather than of its terms.

Facts & Assumptions

Given: The alternating sequence $(s_{k})$ with $s_{0} = 1$ and $s_{σ (k)} = - s_{k}$ , its partial sums $S_{n} = \sum_{k < n} s_{k}$ , and its Cesaro means $σ_{n} = (n + 1)^{- 1} S_{n + 1}$ .

[L1]

The alternating sequence and its index maps $e$ (even indices) and $o$ (odd indices), with $N$ the disjoint union of their ranges and $s_{e_{j}} = 1$ , $s_{o_{j}} = - 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Its partial sums satisfy $S_{e_{j}} = 0$ and $S_{o_{j}} = 1$ , and consequently $∣ σ_{n} ∣ \leq (n + 1)^{- 1}$ and $σ_{n} \to 0$ ; this is proved in FALSE: if the Cesaro means of a sequence converge then the sequence converges, steps 2.1, 3.1, 4.1 and 5.1 there.

[L3]

$(s_{k})$ is bounded and does not converge (FALSE: every bounded sequence converges).

[L4]

The Cesaro means and $(C, 1)$ -summability (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability, Finite sums and finite products, by recursion, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L5]

Convergence of a real sequence (Limits and Cauchy sequences of reals); the reciprocal Archimedean property (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

Order arithmetic: $(n + 1) \cdot 1_{R} > 0$ (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 $u \geq 0$ (Basic properties of the absolute value); the order is total (Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1

$S_{m} = 0$ when $m$ is even and $S_{m} = 1$ when $m$ is odd, since $N$ is the disjoint union of the ranges of $e$ and $o$ and $S_{e_{j}} = 0$ , $S_{o_{j}} = 1$ .

L1L2

1.2

$(s_{k})$ does not converge.

2.1

Hence $σ_{n} = (n + 1)^{- 1} S_{n + 1}$ equals $(n + 1)^{- 1}$ when $n$ is even, because $n + 1$ is then odd, and equals $0$ when $n$ is odd; in particular $σ_{0} = 1$ , $σ_{1} = 0$ , $σ_{2} = 1 / 3$ , $σ_{3} = 0$ , $σ_{4} = 1 / 5$ and $σ_{5} = 0$ .

step 1.1L4L6

2.2

$∣ σ_{n} ∣ \leq (n + 1)^{- 1}$ for every $n$ , and given a real $ε > 0$ a natural $m \geq 1$ with $1 / m < ε$ gives $∣ σ_{n} ∣ \leq (n + 1)^{- 1} < ε$ for all $n \geq m$ ; so $\lim_{n} σ_{n} = 0$ .

step 1.1L2L5L6

3.1

$(s_{k})$ is therefore $(C, 1)$ -summable to $0$ and divergent.

step 1.2step 2.1step 2.2L4∎

Remarks

The means converge but are not monotone, and they are not even eventually of one shape: they alternate between $0$ and a positive value shrinking like $1 / (n + 1)$ . Convergence of a Cesaro transform therefore carries no monotonicity information, which is another way of seeing that the transform loses the oscillation rather than damping it.
Where the $1 / 2$ comes from. The classical assertion " $1 - 1 + 1 - 1 + \dots = 1 / 2$ " is about the partial sums $S_{m}$ , which are $0, 1, 0, 1, \dots$ ; their Cesaro means tend to $1 / 2$ . This library has no theory of series yet, so nothing above asserts it; the sequence averaged here is $(s_{k})$ itself, whose means tend to $0$ .
This is not a failure of the Cesaro matrix. That matrix is regular (The Cesaro matrix satisfies the Silverman-Toeplitz conditions, giving a second proof of the Cesaro mean theorem): it never changes a limit that exists. What it does here is assign a value where no limit exists, which is exactly what a summability method is for.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Cesàro summation (Wikipedia) (standard reference, not scraped)
Grandi's series (Wikipedia) (standard reference, not scraped)

The Cesaro means of (−1)k(−1)k converge to 00 although the sequence diverges