# Cesàro summation

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In mathematical analysis, Cesàro summation (also known as the Cesàro mean[1][2]) assigns values to some infinite sums that are not necessarily convergent in the usual sense. The Cesàro sum is defined as the limit, as n tends to infinity, of the sequence of arithmetic means of the first n partial sums of the series.

This special case of a matrix summability method is named for the Italian analyst Ernesto Cesàro (1859–1906).

The term summation can be misleading, as some statements and proofs regarding Cesàro summation can be said to implicate the Eilenberg–Mazur swindle. For example, it is commonly applied to Grandi's series with the conclusion that the sum of that series is 1/2.

## Definition

Let ${\displaystyle (a_{n})_{n=1}^{\infty }}$ be a sequence, and let

${\displaystyle s_{k}=a_{1}+\cdots +a_{k}=\sum _{n=1}^{k}a_{n}}$

be its kth partial sum.

The sequence (an) is called Cesàro summable, with Cesàro sum A${\displaystyle \mathbb {R} }$, if, as n tends to infinity, the arithmetic mean of its first n partial sums s1, s2, ..., sn tends to A:

${\displaystyle \lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}s_{k}=A.}$

The value of the resulting limit is called the Cesàro sum of the series ${\displaystyle \textstyle \sum _{n=1}^{\infty }a_{n}.}$ If this series is convergent, then it is Cesàro summable and its Cesàro sum is the usual sum.

## Examples

### First example

Let an = (−1)n for n ≥ 0. That is, ${\displaystyle (a_{n})_{n=0}^{\infty }}$ is the sequence

${\displaystyle (1,-1,1,-1,\ldots ).}$

Let G denote the series

${\displaystyle G=\sum _{n=0}^{\infty }a_{n}=1-1+1-1+1-\cdots }$

The series G is known as Grandi's series.

Let ${\displaystyle (s_{k})_{k=0}^{\infty }}$ denote the sequence of partial sums of G:

{\displaystyle {\begin{aligned}s_{k}&=\sum _{n=0}^{k}a_{n}\\(s_{k})&=(1,0,1,0,\ldots ).\end{aligned}}}

This sequence of partial sums does not converge, so the series G is divergent. However, G is Cesàro summable. Let ${\displaystyle (t_{n})_{n=1}^{\infty }}$ be the sequence of arithmetic means of the first n partial sums:

{\displaystyle {\begin{aligned}t_{n}&={\frac {1}{n}}\sum _{k=0}^{n-1}s_{k}\\(t_{n})&=\left({\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 \right).\end{aligned}}}

Then

${\displaystyle \lim _{n\to \infty }t_{n}=1/2,}$

and therefore, the Cesàro sum of the series G is 1/2.

### Second example

As another example, let an = n for n ≥ 1. That is, ${\displaystyle (a_{n})_{n=1}^{\infty }}$ is the sequence

${\displaystyle (1,2,3,4,\ldots ).}$

Let G now denote the series

${\displaystyle G=\sum _{n=1}^{\infty }a_{n}=1+2+3+4+\cdots }$

Then the sequence of partial sums ${\displaystyle (s_{k})_{k=1}^{\infty }}$ is

${\displaystyle (1,3,6,10,\ldots ).}$

Since the sequence of partial sums grows without bound, the series G diverges to infinity. The sequence (tn) of means of partial sums of G is

${\displaystyle \left({\frac {1}{1}},{\frac {4}{2}},{\frac {10}{3}},{\frac {20}{4}},\ldots \right).}$

This sequence diverges to infinity as well, so G is not Cesàro summable. In fact, for any sequence 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

{\displaystyle {\begin{aligned}A_{n}^{-1}&=a_{n}\\A_{n}^{\alpha }&=\sum _{k=0}^{n}A_{k}^{\alpha -1}\end{aligned}}}

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

${\displaystyle (\mathrm {C} ,\alpha ){\text{-}}\sum _{j=0}^{\infty }a_{j}=\lim _{n\to \infty }{\frac {A_{n}^{\alpha }}{E_{n}^{\alpha }}}}$

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

${\displaystyle (\mathrm {C} ,\alpha ){\text{-}}\sum _{j=0}^{\infty }a_{j}=\lim _{n\to \infty }\sum _{j=0}^{n}{\frac {\binom {n}{j}}{\binom {n+\alpha }{j}}}a_{j}.}$

Even more generally, for α${\displaystyle \mathbb {R} }$ \ ${\displaystyle \mathbb {Z} }$, let Aα
n
be implicitly given by the coefficients of the series

${\displaystyle \sum _{n=0}^{\infty }A_{n}^{\alpha }x^{n}={\frac {\displaystyle {\sum _{n=0}^{\infty }a_{n}x^{n}}}{(1-x)^{1+\alpha }}},}$

and Eα
n
as above. In particular, Eα
n
are the binomial coefficients of power −1 − α. Then the (C, α) sum of Σan 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 ${\displaystyle \textstyle \int _{0}^{\infty }f(x)\,dx}$ is (C, α) summable if

${\displaystyle \lim _{\lambda \to \infty }\int _{0}^{\lambda }\left(1-{\frac {x}{\lambda }}\right)^{\alpha }f(x)\,dx}$

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

${\displaystyle \lim _{\lambda \to \infty }{\frac {1}{\lambda }}\int _{0}^{\lambda }\int _{0}^{x}f(y)\,dy\,dx}$

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.