# The Cesàro Summability of a Series

We will now take a break from Fourier series for a moment to discuss a special type of summability known as Cesàro summability which will be useful when we delve deeper into the theory on Fourier series. We begin by defining what it means for a series of real numbers to be Cesàro summable.

Definition: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a series of real numbers and let $(s_n)_{n=1}^{\infty}$ denote the corresponding sequence of partial sums where for each $n \in \mathbb{N}$, $\displaystyle{s_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + ... + a_n}$. For each $n \in \mathbb{N}$ let $\displaystyle{\sigma_n = \frac{1}{n} \sum_{k=1}^{n} s_k = \frac{s_1 + s_2 + ... + s_n}{n}}$. Then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to be Cesàro Summable or $(C, 1)$ Summable if $(\sigma_n)_{n=1}^{\infty}$ converges to some $S \in \mathbb{R}$ for which $S$ is called the Cesàro Sum of $\displaystyle{\sum_{n=1}^{\infty} a_n}$ written $\displaystyle{\sum_{n=1}^{\infty} a_n = S \: \: (C, 1)}$. |

It is very important to note that a divergent series of real numbers may be Cesàro summable. For example, consider the following series:

(1)This series diverges since $\displaystyle{\lim_{n \to \infty} (-1)^{n+1} \neq 0}$. Now, the sequence of partial sums for this series is:

(2)Therefore the values of $\sigma_n$ are:

(3)Clearly $\displaystyle{\lim_{n \to \infty} \sigma_n = \frac{1}{2}}$ and so $\displaystyle{\sum_{n=1}^{\infty} (-1)^{n+1}}$ is Cesàro summable to which we write:

(4)We will now look at a very important theorem which relates the convergence of a series of real numbers with Cesàro summability.

Theorem 1: If $\displaystyle{\sum_{n=1}^{\infty} a_n = S}$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is Cesàro summable and $\displaystyle{\sum_{n=1}^{\infty} a_n = S \: \: (C, 1)}$. |

**Proof:**Suppose that $\displaystyle{\sum_{n=1}^{\infty} a_n = S}$, and let $(s_n)_{n=1}^{\infty}$ denote the sequence of partial sums to this series. Then $\displaystyle{\lim_{n \to \infty} s_n = S}$. So for $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

- Now:

- And as $n \to \infty$ the righthand side of the inequality above goes to $\epsilon$, so $\mid \sigma_n - S \mid < \epsilon$ for all $\epsilon > 0$. Thus $\displaystyle{\lim_{n \to \infty} \sigma_n = S}$ which shows that: