# Cesàro summation

For the song "Cesaro Summability" by the band Tool, see Ænima.

In mathematical analysis, Cesàro summation assigns values to some infinite sums that are not convergent in the usual sense, while coinciding with the standard sum if they are convergent. The Cesàro sum is defined as the limit of the arithmetic mean of the partial sums of the series.

Cesàro summation is named for the Italian analyst Ernesto Cesàro (1859–1906).

## Definition

Let {an} be a sequence, and let

$s_k = a_1 + \cdots + a_k$

be the kth partial sum of the series

$\sum_{n=1}^\infty a_n.$

The series $\sum_{n=1}^\infty a_n$ is called Cesàro summable, with Cesàro sum $A \in \R$, if the average value of its partial sums $s_k$ tends to $A$:

$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n s_k = A.$

In other words, the Cesàro sum of an infinite series is the limit of the arithmetic mean (average) of the first n partial sums of the series, as n goes to infinity. It is easy to show that any convergent series is Cesàro summable, and the sum of the series agrees with its Cesàro sum. However, as the first example below demonstrates, there are series that diverge but are nonetheless Cesàro summable.

## Examples

Let an = (−1)n+1 for n ≥ 1. That is, {an} is the sequence

$1, -1, 1, -1, \ldots.\,$

and let G denote the series $\sum_{n=1}^\infty a_n =1-1+1-1+1-\cdots$

Then the sequence of partial sums {sn} $= \sum_{k=1}^n a_k$ is

$1, 0, 1, 0, \ldots,\,$

so that the series G, known as Grandi's series, clearly does not converge. On the other hand, the terms of the sequence {tn} of the (partial) means of the {sn} where

$t_n = \frac{1}{n}\sum_{k=1}^n s_k$

are

$\frac{1}{1}, \,\frac{1}{2}, \,\frac{2}{3}, \,\frac{2}{4}, \,\frac{3}{5}, \,\frac{3}{6}, \,\frac{4}{7}, \,\frac{4}{8}, \,\ldots,$

so that

$\lim_{n\to\infty} t_n = 1/2.$

Therefore the Cesàro sum of the series G is 1/2.

On the other hand, now let an = n for n ≥ 1. That is, {an} is the sequence

$1, 2, 3, 4, \ldots.\,$

and let G now denote the series $\sum_{n=1}^\infty a_n =1+2+3+4+5+\cdots$

Then the sequence of partial sums {sn} is

$1, 3, 6, 10, \ldots,\,$

and the evaluation of G diverges to infinity. The terms of the sequence of means of partial sums {tn } are here

$\frac{1}{1}, \,\frac{4}{2}, \,\frac{10}{3}, \,\frac{20}{4}, \,\ldots.$

Thus, this sequence diverges to infinity as well as G, and G is now not Cesàro summable. In fact, any series which diverges to (positive or negative) infinity the Cesàro method also leads to a sequence that diverges likewise, and hence such a series is not Cesàro summable.

## (C, α) summation

In 1890, Ernesto Cesàro stated a broader family of summation methods which have since been called (C, α) for non-negative integers α. The (C, 0) method is just ordinary summation, and (C, 1) is Cesàro summation as described above.

The higher-order methods can be described as follows: given a series Σan, define the quantities

$A_n^{-1}=a_n;\quad A_n^\alpha=\sum_{k=0}^n A_k^{\alpha-1}$

(where the upper indices do not denote exponents) and define Enα to be Anα for the series 1 + 0 + 0 + 0 + · · ·. Then the (C, α) sum of Σan is denoted by (C, α)-Σan and has the value

$(C,\alpha)-\sum_{j=0}^\infty a_j=\lim_{n\to\infty}\frac{A_n^\alpha}{E_n^\alpha}$

if it exists (Shawyer & Watson 1994, pp.16-17). This description represents an $\alpha$-times iterated application of the initial summation method and can be restated as

$(C,\alpha)-\sum_{j=0}^\infty a_j = \lim_{n\to\infty} \sum_{j=0}^n \frac{{n \choose j}}{{n+\alpha \choose j}} a_j.$

Even more generally, for $\alpha\in\mathbb{R}\setminus(-\mathbb{N})$, let Anα be implicitly given by the coefficients of the series

$\sum_{n=0}^\infty A_n^\alpha x^n=\frac{\displaystyle{\sum_{n=0}^\infty a_nx^n}}{(1-x)^{1+\alpha}},$

and Enα as above. In particular, Enα are the binomial coefficients of power −1 − α. Then the (C, α) sum of Σ an is defined as above.

If Σan has a (C, α) sum, then it also has a (C, β) sum for every β>α, and the sums agree; furthermore we have an = o(nα) if α > −1 (see little-o notation).

## Cesàro summability of an integral

Let α ≥ 0. The integral $\scriptstyle{\int_0^\infty f(x)\,dx}$ is Cesàro summable (C, α) if

$\lim_{\lambda\to\infty}\int_0^\lambda\left(1-\frac{x}{\lambda}\right)^\alpha f(x)\, dx$

exists and is finite (Titchmarsh 1948, §1.15). The value of this limit, should it exist, is the (C, α) sum of the integral. Analogously to the case of the sum of a series, if α=0, the result is convergence of the improper integral. In the case α=1, (C, 1) convergence is equivalent to the existence of the limit

$\lim_{\lambda\to \infty}\frac{1}{\lambda}\int_0^\lambda\left\{\int_0^xf(y)\, dy\right\}\,dx$

which is the limit of means of the partial integrals.

As is the case with series, if an integral is (C,α) summable for some value of α ≥ 0, then it is also (C,β) summable for all β > α, and the value of the resulting limit is the same.