Cesàro summation

Examples

First example

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

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

Let G denote the series

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

The series G is known as Grandi's series.

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

${\displaystyle \begin{array}{cc}{s}_k& =\sum _{n=0}^k{a}_n\\ (s_k)& =(1,0,1,0,\dots ).\end{array}}$

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

${\displaystyle \begin{array}{cc}{t}_n& =\frac{1}{n}\sum _{k=0}^{n-1}{s}_k\\ (t_n)& =\left(\frac{1}{1},\frac{1}{2},\frac{2}{3},\frac{2}{4},\frac{3}{5},\frac{3}{6},\frac{4}{7},\frac{4}{8},\dots \right).\end{array}}$

Then

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{t}_n=1/2,}$

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

Second example

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

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

Let G now denote the series

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

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

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

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

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

This sequence diverges to infinity as well, so G is not Cesàro summable. In fact, for the series of any sequence which diverges to (positive or negative) infinity, the Cesàro method also leads to the series of 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 Σa_n, define the quantities

${\displaystyle \begin{array}{cc}{A}_n^{-1}& =a_n\\ A_n^{\alpha }& =\sum _{k=0}^n{A}_k^{\alpha -1}\end{array}}$

(where the upper indices do not denote exponents) and define E^α
_n to be A^α
_n for the series 1 + 0 + 0 + 0 + .... Then the (C, α) sum of Σa_n is denoted by (C, α)-Σa_n and has the value

${\displaystyle (\mathrm{C},\alpha )\text{-}\sum _{j=0}^{\mathrm{\infty }}{a}_j=\underset{n\to \mathrm{\infty }}{lim}\frac{A_n^{\alpha }}{E_n^{\alpha }}}$

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

${\displaystyle \begin{array}{cc}(\mathrm{C},\alpha )\text{-}\sum _{j=0}^{\mathrm{\infty }}{a}_j& =\underset{n\to \mathrm{\infty }}{lim}\sum _{j=0}^n\frac{(\genfrac{}{}{0ex}{}{n}{j})}{(\genfrac{}{}{0ex}{}{n+\alpha }{j})}{a}_j\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{j=0}^n\frac{{\left(n-j+1\right)}_{\alpha }}{{\left(n+1\right)}_{\alpha }}{a}_j\text{.}\end{array}}$

Even more generally, for α ∈ ${\displaystyle \mathbb{R}}$ \ ${\displaystyle \mathbb{Z}}$⁻, let A^α
_n be implicitly given by the coefficients of the series

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{A}_n^{\alpha }{x}^n=\frac{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n}}{(1-x{)}^{1+\alpha }},}$

and E^α
_n as above. In particular, E^α
_n are the binomial coefficients of power −1 − α. Then the (C, α) sum of Σa_n is defined as above.

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

Cesàro summability of an integral

Let α ≥ 0. The integral ${\displaystyle {\int }_0^{\mathrm{\infty }}f(x)dx}$ is (C, α) summable if

${\displaystyle \underset{\lambda \to \mathrm{\infty }}{lim}{\int }_0^{\lambda }{\left(1-\frac{x}{\lambda }\right)}^{\alpha }f(x)dx}$

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

${\displaystyle \underset{\lambda \to \mathrm{\infty }}{lim}\frac{1}{\lambda }{\int }_0^{\lambda }{\int }_0^{x}f(y)dydx}$

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.