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?
 A: 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.
A: I think the problem with your strategy appears in your first step, when you are trying to approximate $A - \frac{1}{n}z_1 - \cdots - \frac{1}{n}z_n$. The problem is that you really aren't ever considering the difference between $A$ and $z_n$ when you are coming up with your bounds; remember, the only information you have is that $z_n\to A$, so you have to use that the difference between $A$ and $z_n$ goes to $0$. I think it would be more fruitful to instead first rewrite as $$
A - \frac{z_1}{n} - \cdots - \frac{z_n}{n} = \frac{(A-z_1)}{n} + \cdots + \frac{(A-z_n)}{n},$$ and then apply the triangle inequality. Maybe you can finish the argument from there?
