Showing that $\lim\limits_{n \to\infty} z_n = A$ implies $\lim\limits_{n \to\infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A$

Question

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:

1. 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

   $$ \left| A - s_n \right| \le \left| A - \frac{z_n}{n} \right| + \left| - \frac{z_{n-1}}{n} - \ldots - \frac{z_{1}}{n} \right| \le \left| A - \frac{z_n}{n} \right| + \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| $$

2. 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 $0$.

3. 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 $\left| A - \frac{z_n}{n} \right| \rightarrow \left| A - 0 \right| = \left| A \right|$.

4. 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?

Answer 1

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.

Answer 2

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?