The Cesàro Summability of a Series

# 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)
\begin{align} \quad (s_n)_{n=1}^{\infty} = (1, 0, 1, 0, 1, 0, ...) \end{align}

Therefore the values of $\sigma_n$ are:

(3)
\begin{align} \quad (\sigma_n)_{n=1}^{\infty} = \left ( 1, \frac{1}{2}, \frac{2}{3}, \frac{2}{4}, \frac{3}{5}, \frac{3}{6}, ... \right ) \end{align}

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)
\begin{align} \quad \sum_{n=1}^{\infty} (-1)^{n+1} = \frac{1}{2} \: \: (C, 1) \end{align}

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:
(5)
\begin{align} \quad \mid s_n - S \mid < \epsilon \end{align}
• Now:
(6)
\begin{align} \quad \mid \sigma_n - S \mid & = \biggr \lvert \frac{s_1 + s_2 + ... + s_n}{n} - S \biggr \rvert \\ &= \biggr \lvert \frac{(s_1 - S) + (s_2 - S) + ... + (s_n - S)}{n} \biggr \rvert \\ &\leq \frac{\mid s_1 - S \mid + \mid s_2 - S \mid + ... + \mid s_{N-1} - S \mid}{n} + \frac{\mid s_{N} - S \mid + \mid s_{N+1} - S \mid + ... + \mid s_n - S \mid}{n} \\ & < \frac{\mid s_1 - S \mid + \mid s_2 - S \mid + ... + \mid s_{N-1} - S \mid}{n} + \frac{(n - N + 1)\epsilon}{n} \end{align}
• 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:
(7)
\begin{align} \quad \sum_{n=1}^{\infty} a_n = S \: \: (C, 1) \quad \blacksquare \end{align}