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

Let $(s_k)$ be the alternating sequence 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 $s_{\sigma(k)} = -s_k$, usually written $s_k = (-1)^k$. Its Cesaro means (The Cesaro means $\sigma_n = (x_0 + \dots + x_n)/(n+1)$ and $(C,1)$-summability) are

$\sigma_n \;=\; \frac{s_0 + \dots + s_n}{n+1} \;=\; \begin{cases} \dfrac{1}{n+1} & n \text{ even},\\[4pt] 0 & n \text{ odd},\end{cases}$

so the first few values are

$\sigma_0 = 1,\quad \sigma_1 = 0,\quad \sigma_2 = \tfrac13,\quad \sigma_3 = 0,\quad \sigma_4 = \tfrac15,\quad \sigma_5 = 0,\ \dots$

and $\lim_n \sigma_n = 0$, 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 $\tfrac12$, that being the Cesaro limit of its partial sums rather than of its terms.

• 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.