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For $a_n=(-1{)}^n(n+1)$, direct finite summation gives $s_{2m}=m+1$ and $s_{2m+1}=-(m+1)$. ConsequentlyThe distinct even and odd subsequential limits prevent first-order Cesàro summation. More averaging is a different method; a value assigned by another method is not a first-order Cesàro sum.

Construct partial sums $s_n$ from alternating blocks of zero and one, with lengths $1,3,9,27,\dots$. 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 $n\ge 1$; 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 isThese 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.

The Cesàro mean of the first $n\ge 1$ 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 $p\ge 1$ 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.

The finite limit of a sequence of averaged partial sums in the declared Cesàro summation convention, when that limit exists. It can exist when an ordinary sum does not, but every ordinarily convergent series has the same Cesàro sum as its ordinary sum.

For the geometric series with terms $z^n$, $z\ne 1$, the partial sums are $(1-z^{n+1})/(1-z)$. Averaging the finite expressions givesIf $|z|\le 1$ and $z\ne 1$, the second term has absolute value at most $2/((N+1)|1-z{|}^2)$, and therefore tends to zero. On the unit circle, except at $z=1$, this yields a first-order Cesàro sum even though the terms do not tend to zero. For $z=1$, instead $s_n=n+1$ and ${\sigma }_N=(N+2)/2$, so there is no finite Cesàro sum.^[ref]