__Definition__The Cesàro mean of a sequence \( a_ n \) is defined as the limit

\begin{equation*}

\lim_{n \to \infty} b_n \quad\text{with}\quad b_n = \frac{1}{n} \left( a_1 + \cdots + a_n \right)

\end{equation*}

__Abelian Property__If the limit of the sequence \( a_ n \) exists, then the Cesàro mean also exists, and is equal to this limit:

If \(\displaystyle \lim_{n \to \infty} a_n =a \) exists, then \( \displaystyle\lim_{n \to \infty} \frac{1}{n} \left( a_1 + \cdots + a_n \right) = a \)

The proof is very short if one uses Landau's \( o \) notation: one has

\begin{equation*}b_n - a = \frac{1}{n} \sum_{k = 1}^n \left(a_k - a \right)

\end{equation*}

Because \( \displaystyle\lim_{k \to \infty} a_k =a \) means \( a_k = a + o(1) \), one has thus

\begin{equation*}

b_n - a = \frac{1}{n} \sum_{k = 1}^n o(1)

\end{equation*}

Because \( \sum_{k = 1}^n o(1) = o(n) \) and \( \frac{1}{n} o(n) = o(1) \), it follow that \( b_n -a = o(1) \). This means \( \displaystyle\lim_{n \to \infty} b_n =a \) . QED

__Tauberian Theorem__Conversily, there is the following Tauberian theorem

If \( \displaystyle\lim_{n \to \infty} \frac{1}{n} \left( a_1 + \cdots + a_n \right) = a \) exists, and \( \Delta a_n = o\left( \frac{1} {n} \right) \) , then \(\displaystyle \lim_{n \to \infty} a_n =a \)

Proof: Partial summation gives

\begin{equation*}

a_n - b_n = \frac{1}{n} \sum_{k=1}^{n-1} \Delta a_k \cdot k

\end{equation*}

because \( \Delta a_k = o\left( \frac{1} {k} \right) \) and \( o\left( \frac{1} {k} \right) \cdot k = o(1) \) we have

\begin{equation*}

a_n - b_n = \frac{o(n)}{n} = o(1)

\end{equation*}

This means \( \displaystyle\lim_{n \to \infty} ( a_n - b_n ) = 0 \) thus \( \displaystyle\lim_{n \to \infty} a_n =\displaystyle\lim_{n \to \infty} b_n = a \). QED

Remark:

- The condition \( \Delta a_n = o\left( \frac{1} {n} \right) \) is a Tauberian condition, it can be weakened to \( \Delta a_n = O\left( \frac{1} {n} \right) \)
- The result can also be translated to a Tauberian theorem about Cesàro summability:

If \( \displaystyle\sum_{n = 0}^{\infty} c_n \) is Cesàro summable to \( s \) and \( c_n = o\left( \frac{1} {n} \right) \) , then \( \displaystyle\sum_{n = 0}^{\infty} c_n =s \)

## No comments:

## Post a Comment