Tauberian Theorem for Cesàro mean

Because the proof of the prime number theorem is related to Tauberian theorems of various kinds (see for example this paper by Mueger), I decided to prove a Tauberian theorem for a very simple case, namely the case of Cesàro means. It seems difficult to find the proof online, I therefore write one down here.



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 $\displaystyle \lim_{n \to \infty} a_n =a$

Remark: I use the notation $\Delta a_n = a_{n + 1} - a_n$
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 $\displaystyle\sum_{n = 0}^{\infty} c_n =s$