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

Prove that if $\lim_{n \to \infty}z_{n}=A$ then: $$\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A$$ I was thinking spliting it in: $$(z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})$$ where $N$ is value of $n$ for which $|A-z_{n}|<\epsilon$ then taking the limit of this sum devided by $n$ , and noting that the second sum is as close as you wish to $nA$ while the first is as close as you wish to $0$. Not sure if this helps....

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 $