We define:
series $\sum_{n=1}^{\infty} a_n$
$s_n=\sum_{k=1}^{n} a_k, n\in \mathbb{N}.$
$\sigma_n=\frac{1}{n}\sum_{k=1}^n s_k$
$s=\lim_{n\to \infty} s_n$
Proposition: If sequence $(s_n)_n$ converges and if $s=\lim_{n\to \infty} s_n$ then sequence $(\sigma_n)_n$ also converges and $s=\lim_{n\to \infty} \sigma_n$.
Proof: Let $s=\lim_{n\to \infty} s_n$ and $\epsilon \gt0$. Let $\epsilon_n=s_n-s,n\in \mathbb{N}$.There exists $m\in \mathbb{N}$ so that $\forall n\in \mathbb{N}, (n\gt m) \Rightarrow (|\epsilon_n \lt \frac{\epsilon}{3}|)$. Now for $n \gt m+1$ we have $$\sigma_n=\frac{s_1+…+s_{m+1}}{n}+\frac{s_{m+2}+…+s_n}{n} =$$ $$\frac{s_1+…+s_{m+1}}{n}+\frac{n-m-1}{n}s+\frac{\epsilon_{m+2}+…+\epsilon_n}{n}$$
From that we have (and I don't understand why)$$|\sigma_n-s|\lt \frac{|s_1+…+s_n|}{n}+\frac{m+1}{n}|s|+\frac{n-m-1}{n}\frac{\epsilon}{3}$$
Lets take $n_{\epsilon} \gt m+1$ so that $\frac{|s_1+…+s_n|}{n_{\epsilon}}\lt \frac{\epsilon}{3}$ and $\frac{(m+1)|s|}{n_{\epsilon}} \lt \frac{\epsilon}{3}$.
Now, $\forall n\in \mathbb{N}, (n \gt n_{\epsilon}) \Rightarrow |\sigma_n-s|\lt \epsilon$
Q.E.D
If someone could explain that transition, I'd be happy.
Best Answer
Subtract $s$ from $\sigma_n$, take absolute values, and apply the triangle inequality:
$$\begin{align*} \left|\sigma_n-s\right|&=\left|\frac{s_1+\ldots+s_{m+1}}n-\frac{m+1}ns+\frac{\epsilon_{m+2}+\ldots+\epsilon_n}n\right|\\ &\le\frac{|s_1+\ldots+s_{m+1}|}n+\frac{m+1}n|s|+\frac{|\epsilon_{m+2}|+\ldots+|\epsilon_n|}n\\ &<\frac{|s_1+\ldots+s_{m+1}|}n+\frac{m+1}n|s|+\frac{n-m-1}n\cdot\frac{\epsilon}3\;. \end{align*}$$
The final step uses the fact that each $|\epsilon_k|<\frac{\epsilon}3$, and there are $n-m-1$ of these $\epsilon_k$ terms in the numerator.