Cesàro mean ($\frac{1}{n}{\sum }_{k=1}^n{a}_k$)

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.

For a formal cosine series ${\sum }_j{a}_j\cos (j!t)$ with uniformly bounded coefficients, the Fourier Cesàro mean at $N_J$ differs from its ordinary partial sum by $-(N_J+1{)}^{-1}{\sum }_{j\le J}j!a_j\cos (j!t)$. Since ${\sum }_{j\le J}j!\le 2J!$, the difference is $O(1/J)$ uniformly in $t$. Thus any divergent behavior of these ordinary cutoff sums persists along the same Cesàro subsequence. Averaging by the sparse coefficient index is a different summation operation from averaging the actual Fourier frequency cutoffs.

Let $C{a}_n=n^{-1}{\sum }_{j=1}^n{a}_j$ be the Cesàro mean of a sequence. If $|a_n|\le Cn$ for every positive integer $n$, the triangle inequality gives $|C{a}_n|\le C(n+1)/2\le Cn$. Mathematical induction repeats this estimate with the same constant for every iterate. An eventual $O(n)$ bound can be made global by increasing $C$ to absorb the finitely many initial terms. Consequently iterated Cesàro means preserve linear growth in Big O notation; this bound alone does not assert convergence or summability.

If a real or complex sequence converges to $a$, its Cesàro means have the same limit of a sequence. This is why Cesàro summation is a regular summation method when applied to the partial sums of a series.

Proof outline. Given $\epsilon >0$, choose $K$ so that $|a_k-a|<\epsilon$ for $k>K$. The triangle inequality gives, for $n>K$,The finite initial contribution tends to zero, and the remaining contribution is at most $\epsilon$. The converse fails: the alternating sequence $(-1{)}^n$ has Cesàro means tending to zero while its terms do not converge.