= Bounded partial sums need not be Cesàro summable
Construct <partial sums> $s_n$ from alternating blocks of zero and one, with lengths $1,3,9,27,\ldots$. Block $j$, beginning with $j=0$, has length $3^j$ and value zero when $j$ is <even>, one when $j$ is <odd>. Define a <series> by $a_0=s_0$ and $a_n=s_n-s_{n-1}$ for $n\geq1$; the <telescoping series> identity recovers these chosen <partial sums>. They remain in $[0,1]$.
At the end of odd block $2m+1$, the <Cesàro mean> is exactly $3/4$. At the end of even block $2m$, it is
$$
\frac{3(9^m-1)}{4(3\cdot9^m-1)}\longrightarrow\frac14.
$$
These two <subsequences> have different <subsequential limits>, so the <series> has no first-order <Cesàro sum>. A later block remains a substantial fraction of everything averaged so far. Boundedness alone therefore does not ensure a stable average.
\Image[/cesaro-means-periodic-and-long-blocks.png]
{title=Periodic <partial sums> have a stable average, while longer alternating blocks prevent a <Cesàro sum>}
{description=Original exact constructed <sequences>. In the left panel, $s_n$ repeats $0,2,1$, giving <Cesàro means> tending to one. In the right panel, the <partial sums> alternate between blocks of zero and one whose lengths are powers of three. The horizontal axis uses the number $N+1$ of included <partial sums>.}
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