= Cesàro mean
{c}
{title2=$\frac1n\sum_{k=1}^n a_k$}
= Cesàro average
{c}
{synonym}
The <Cesàro mean> of the first $n\geq1$ terms of a <sequence> is their <arithmetic mean>. It is a cumulative average: increasing $n$ includes the whole initial segment, rather than a fixed-length sliding window. For $a_k=(-1)^k$, the even-length averages are zero and the odd-length averages are $-1/n$, so they tend to zero even though the <sequence> does not converge.
For a <periodic sequence> with period $p\geq1$ and one-cycle <arithmetic mean> $M$, split an initial segment into full cycles and a remainder of fewer than $p$ terms. The full cycles contribute exactly $M$ per term; the remainder contributes a fixed bounded error divided by $n$. Thus the <Cesàro means> tend to $M$. The cycle $0,2,1$ in the <bounded partial sums need not be Cesàro summable> illustration has <arithmetic mean> one. Treating it as a sequence of <partial sums> gives a <series> with first-order <Cesàro sum> one; averaging a sequence and summing its terms remain different operations.
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