Convergence of sequence of partial sums using convergence of term This is Exercise 2 on page 37 of Complex Analysis by Ahlfors.  Sorry if this is an easy question. I'm slowly teaching myself using this book, but I cannot seem to figure this one out with so little information.
If $$\lim_{n\to\infty}z_n=A,$$ show that $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nz_n=A.$$
Where I am is not much further, but I proceeded by essentially defining $$\epsilon_m\equiv\text{max}(|A-z_{n\geq m}|)$$ and showing that the sequence in question is bounded: $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nz_n\geq A-\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\epsilon_k$$ and $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nz_n\leq A+\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\epsilon_k.$$ With the partial sums in the limits on the right hand side, $$s_n\equiv \sum_{k=1}^n\epsilon_k,$$ I know that if I can show that $s_n$ converges, then the bounds are contracted and my proof is complete.  However, all I know about $s_n$ is that it is bounded from below and monotonically increasing since ${\epsilon_k}$ is a monotonically decreasing sequence with range [0,$\epsilon_1$].  Am I on the right track and can I show the $s_n$ is bounded above and thus converges?  If so, how might I proceed?
I have tried searching here and elsewhere, but worked solutions of Ahlfors's text seem to always skip this problem or I am too new at this material to know how to properly search for it.  I will gladly accept a simple link to a worked solution or a good hint.
 A: There is perhaps an easier way to do it without introducing the $\epsilon_m$'s : Consider
$$
\left | \frac{1}{n}\sum_{k=1}^n z_n - A\right | \leq \frac{1}{n}\sum_{k=1}^n |z_n - A| 
\quad\quad\text{(1)}
$$
So for $\epsilon > 0$, choose $N_0 \in \mathbb{N}$ such that
$$
|z_k - A| < \epsilon/2 \quad\forall k\geq N_0
$$
Now let $M := \max_{k \leq N_0}\{|z_k - A|\}$, and choose $N_1 \in \mathbb{N}$ such that $N_1 \geq N_0$ and
$$
\frac{M}{n} < \epsilon/2 \quad\forall n\geq N_1
$$
Now can you try making the RHS of $(1)$ smaller than $\epsilon$?
A: Let $\epsilon>0$, and choose $N$ so large that almost all of the terms $z_1$ up to $z_N$ are within $\epsilon$ of $A$. Then show that $$\left|A - \frac{1}{N} \sum_{k=1}^N z_k\right|$$
is small.
You should try to answer the following:


*

*Determine what kind of proportion "almost all" need to be, in terms of $\epsilon$, and in terms of the initial behavior of the sequence.

*Show that you can find an $N$ which realizes the proportion you found in (1).

