Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means

It seems like Homework problem, hence I'll just give hint: $$\frac{z_1+z_2+\cdots +z_n}{n}-A=\frac {(z_1-A)+(z_2-A)+\cdots +(z_n-A)}{n}$$ Now use the defn of limit that for every $\epsilon > 0$ there exists $N_0 \in \mathbb N$ such that $|z_m-A| < \epsilon \ \forall m \geq N_0$

Also remember triangle inequality : $|a_1+a_2+\cdots +a_n| \leq |a_1| + |a_2| +\cdots +|a_n|$

Can you find proper $a_i$ in terms of say $z_i$'s??

Let $\epsilon >0$

We have that $\lim_{n \rightarrow \infty}z_n=A$ thus $\exists n_1 \in \mathbb{N}$ such that $|z_n-A|< \epsilon, \forall n \geqslant n_1$

$|\frac{z_1+...+z_n}{n}-A|=|\frac{(z_1-A)+...(z_{n_1-1}-A)}{n}+\frac{(z_{n_1}-A)+...+(z_n-A)}{n}| \leqslant \frac{|z_1-A|+...+|z_{n_1-1}-A|}{n}+ \frac{|z_{n_1}-A|+...+|z_n-A|}{n}$

Exists $n_2 \in \mathbb{N}$, by Archimedean property, such that $$\frac{|z_1-A|+...+|z_{n_1-1}-A|}{n}< \epsilon, \forall n \geqslant n_2$$

Now for $n \geqslant n_0= \max\{n_1,n_2\}$

$|\frac{z_1+...+z_n}{n}-A| \leqslant \epsilon+ \frac{(n-n_1) }{n}\epsilon<2 \epsilon $

This can be an easy consequence of a more general statement which is from Polya's Problems and Theorems in Analysis:

Let $\{a_n\}_{n=1}^{\infty}$ be a real sequence such that $\lim_{n\to\infty}a_n=a$. And we have a family of finite sequences $\{\{b_{nm}\}_{m=1}^{m=n}\}_{n=1}^{\infty}$: $$ b_{11}\\ b_{21},b_{22}\\ b_{31},b_{32},b_{33}\\ \cdots $$ such that $$ b_{mn}\geq 0 $$ for all $m,n$, and $\sum_{m}b_{nm}=1$ for each $n=1,2,\cdots$. Let $\{c_n\}_{n=1}^{\infty}$ be such that $$ c_n=\sum_{m=1}^na_mb_{nm} $$ Then $\lim_{n\to\infty}c_n=a$ if and only if $\lim_{n\to\infty}b_{nm}=0$ for each $m$.

The question in OP is a special case of the statement by letting $$ b_{nm}=\frac{1}{n},\quad m=1,2,\cdots. $$