if
$$z_n\to 0$$, then $$\frac{1}{n}(z_1+\cdots + z_n)\to 0$$
I'm sorry for asking this one, I've seen it in may places before but I cannot think of a way to solve it, and I don't know how to search it here. Also, I've seen the solution before for real numbers.
Let $\epsilon>0$ be given, and let $N\in\Bbb N$ be such that for all $n>N$ $z_n<\epsilon/2$. Then we have
$${1\over n}\sum_{i=1}^nz_i = {1\over n}(z_1+\ldots + z_N) + {1\over n} (z_{N+1}+\ldots + z_n).$$
Taking absolute values and using the triangle inequality we have
$${1\over n}\left|\sum_{i=1}^nz_i\right| \le {1\over n}\left|z_1+\ldots + z_N\right| + {\epsilon\over 2}.$$
Now if we also choose $n> \displaystyle{2\over\epsilon}N\displaystyle\max_{1\le i\le N}\{|z_i|\}$ we see that this gives us
$${1\over n}\left|\sum_{i=1}^nz_i\right| \le \epsilon$$
as desired.