Showing that $\lim\limits_{n \to\infty} z_n = A$ implies $\lim\limits_{n \to\infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A$ In what follows let all values be in $\mathbb{C}$.  I'm trying to show that if 
$$\lim z_n = A,$$
that then
$$
\lim_{n \to \infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A.
$$
For ease of notation, let $s_n = \frac{1}{n} (z_1 + z_2 + \ldots + z_n)$.
Attempt:


*

*Let $n \in \mathbb{N}$ be arbitrary and consider that
$$
\left| A - s_n \right| = \left|A - \frac{1}{n} (z_1 + \ldots + z_n) \right| = \left|A - \frac{z_1}{n} - \ldots - \frac{z_n}{n} \right| 
$$
so that through repeated applications of the triangle inequality we have that 
$$
\left| A - s_n \right| \le \left| A - \frac{z_n}{n} \right| + \left| - \frac{z_{n-1}}{n} - \ldots - \frac{z_{1}}{n} \right| \le \left| A - \frac{z_n}{n} \right| + \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|
$$

*Now as $n \rightarrow \infty$, we have that the numerator of the term  $\left| -\frac{z_{n-1}}{n} \right|$ approaches $-A$ while the denominators of all of the terms in the sum $\left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|$ approach infinity.  \uline{[Gap]}. Then as $n \rightarrow  \infty$, we have that $\left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|$ approaches $0$.

*On the other hand, we have also that $z_n \rightarrow A$ (by hypothesis) so that the term $\left| A - \frac{z_n}{n} \right|$ can get as close to $\left| A - \frac{A}{n} \right|$ as we'd like.  Yet since $\frac{A}{n} \rightarrow 0$, we have that $\left| A - \frac{z_n}{n} \right| \rightarrow \left| A - 0 \right| = \left| A \right|$.

*Then since
$$
\left( \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| \right) \rightarrow 0
$$
and
$$
\left| A - \frac{z_n}{n} \right| \rightarrow \left| A \right|
$$
we have that 
$$
|A - s_n| \le \left| A - \frac{z_n}{n} \right| + \left( \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| \right) \rightarrow |A| + 0 = |A|.
$$
Question: My argument doesn't quite work since I have shown only that $|A - s_n| \rightarrow |A|$ and yet we want $|A - s_n| \rightarrow |0|$.  Is there a way to keep most of my argument in place and yet actually to prove the desired statement?

In step 1., "through repeated applications of the triangle inequality we have", more interestingly, that
 $$
\left| A - s_n \right|=\left|\frac1n\sum_{k=1}^n(A - z_k) \right| \leqslant\frac1n\sum_{k=1}^n\left| A - z_k  \right| .
$$
Then one can proceed.
Later on, in step 4., you fall prey to the common fallacy that since $\displaystyle\left|\frac{z_k}n\right|\to0$ when $n\to\infty$, for each fixed $k$ (which is quite true), one would be sure that $\displaystyle\sum_{k=1}^{n-1}\left|\frac{z_k}n\right|\to0$. Not so, since thenumber of terms is $n-1$, which is unbounded.
