= First-order Cesàro summation of a geometric series
{title2=$(C,1)\sum_{n=0}^{\infty}z^n=(1-z)^{-1}\quad(|z|\leq1,\ z\neq1)$}
For the <geometric series> with terms $z^n$, $z\neq1$, the <partial sums> are $(1-z^{n+1})/(1-z)$. Averaging the finite expressions gives
$$
\sigma_N=\frac1{1-z}-\frac{z(1-z^{N+1})}{(N+1)(1-z)^2}.
$$
If $|z|\leq1$ and $z\neq1$, 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>.<Cesàro Summation (Mathematica tutorial) by Vladimir Dobrushkin>[Example 2, geometric-series calculation]{ref}
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