Cesàro summation

In mathematical analysis, Cesàro summation (also known as the Cesàro mean or Cesàro limit ) 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 $$(a_n)_{n=1}^\infty$$ be a sequence, and let


 * $$s_k = a_1 + \cdots + a_k= \sum_{n=1}^k a_n$$

be its $k$th partial sum.

The sequence $(a_{n})$ is called Cesàro summable, with Cesàro sum $A ∈ $\mathbb{R}$$, if, as $n$ tends to infinity, the arithmetic mean of its first n partial sums $s_{1}, s_{2}, ..., s_{n}$ tends to $A$:


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

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

First example
Let $a_{n} = (−1)^{n}$ for $n ≥ 0$. That is, $$(a_n)_{n=0}^\infty$$ is the sequence
 * $$(1, -1, 1, -1, \ldots).$$

Let $G$ denote the series
 * $$G = \sum_{n=0}^\infty a_n = 1-1+1-1+1-\cdots $$

The series $G$ is known as Grandi's series.

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

s_k &= \sum_{n=0}^k a_n \\ (s_k) &= (1, 0, 1, 0, \ldots). \end{align}$$

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

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{align}$$ Then
 * $$\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 $a_{n} = n$ for $n ≥ 1$. That is, $$(a_n)_{n=1}^\infty$$ is the sequence


 * $$(1, 2, 3, 4, \ldots).$$

Let $G$ now denote the series


 * $$G = \sum_{n=1}^\infty a_n = 1+2+3+4+\cdots $$

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


 * $$(1, 3, 6, 10, \ldots).$$

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


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

This sequence diverges to infinity as well, so $G$ is 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 $Σa_{n}$, define the quantities


 * $$\begin{align} A_n^{-1}&=a_n \\ A_n^\alpha&=\sum_{k=0}^n A_k^{\alpha-1} \end{align}$$

(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 $Σa_{n}$ is denoted by $(C, α)-Σa_{n}$ and has the value


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

if it exists. This description represents an $α$-times iterated application of the initial summation method and can be restated as


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

Even more generally, for $α ∈ $\mathbb{R}$ \ $\mathbb{Z}$^{−}$, let $Aα n$ 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 $Eα n$ as above. In particular, $Eα n$ 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 $$\textstyle\int_0^\infty f(x)\,dx$$ is $(C, α)$ summable if


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

exists and is finite. 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


 * $$\lim_{\lambda\to \infty}\frac{1}{\lambda}\int_0^\lambda \int_0^x f(y)\, dy\,dx$$

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.