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Absorb the finitely many exceptional initial terms in the Big O notation assumption into a constant $C>0$, so that $|a_n|\le Cn$ for every $n\ge 1$. This is possible because each initial term is finite and each index is positive. We prove the stronger bound $|{}^k{b}_n|\le Cn$ for every $k,n\ge 1$.

The case $k=1$ is the chosen bound. If it holds at $k-1$, the triangle inequality givesMathematical induction completes the proof. Thus ${}^k{b}_n=O(n)$ for every fixed $k$; in fact the same constant works for all the iterated Cesàro means. This is the linear-growth bound for iterated Cesàro means.

A sequence indexed from zero with an integer period $p\ge 1$ satisfying the displayed identity for all $n\ge 0$. Repeating a finite cycle is an example. The Cesàro mean article explains why its cumulative averages approach the one-cycle arithmetic mean.

Define a sequence by $a_n=1$ when $n$ is a power of two, including $1=2^0$, and $a_n=0$ otherwise. Every tail of a sequence contains both a later power of two and a later nonpower. Therefore $\mathrm{liminf}{a}_n=0$ and $\mathrm{limsup}{a}_n=1$, and the sequence does not converge. Its peak indices become increasingly sparse, yet never stop appearing.

Through index $N\ge 1$, the number of peaks is $1+\lfloor {\log }_{2}N\rfloor$, so their fraction isThe set of powers of two has natural density zero. The same fraction is the Cesàro mean of the first $N$ terms, which tends to zero even though the limit superior stays one. At $N=1024$, eleven peaks give mean $11/1024\approx 0.0107422$. An average and a long-run upper edge answer different questions about the same sequence.

A nonprincipal ultrafilter on the natural numbers gives a Banach limit by taking the ultralimit of the Cesàro means of each bounded sequence. Compactness of a closed bounded interval, or disk for complex sequences, supplies the limit. Finite intersections in the filter give linearity; nonnegative averages give positivity; constants give norm one. The Cesàro theorem for convergent sequences gives agreement with ordinary limits because the ultrafilter contains every cofinite set. For the left shift on bounded sequences, the two means differ by $(x_{n+1}-x_1)/n$, which tends to zero, proving shift invariance.

If $x_n\rightharpoonup x$ in a Hilbert space, some subsequence $(x_{n_k})$ has Cesàro means converging in norm to $x$. For $x=0$, choose the subsequence so successive vectors have summably small inner products; expansion of the squared norm of its averages then proves convergence.

Thus Hilbert spaces have the weak form of the Banach-Saks property.