Proof of convergence for reciprocal of convergent sequence

Question

On page - 62 of Bartle and Sherbert's Introduction to Real Analysis, it is proved that-

If $Z=(z_n)$ is a sequence of nonzero numbers that converges to nonzero limit z, then the sequence $(1/z_n)$ of reciprocals converges to $(1/z)$.

However, i found the given proof somewhat difficult to understand, and made my own attempt at what i hoped was a simpler one. The said proof runs as follows-

PROOF:

$(z_n)_n\to z\neq0\quad(n\to\infty)$

$\Rightarrow$ $\forall\epsilon\gt0:\exists N_1\in \mathbb{N}:\forall n\geq N_1:\vert z_n -z \vert\lt\epsilon$

Now, for $(1/z_n)$:

$$\forall n \in \mathbb{N}:\vert 1/z_n -1/z \vert = \vert{ z-z_n\over z_n.z}\vert = { (\vert z_n - z\vert) \over (\vert z_n \vert . \vert z \vert) } \lt {\epsilon \over \vert z_n \vert \vert z \vert } = {\delta \over \vert z_n \vert} \text{, where }\delta ={\epsilon \over \vert z \vert } $$

$\Rightarrow \vert 1/z_n - 1/z \vert \lt \delta / ( \vert z_n \vert)$ , for all $n\in\mathbb{N}$

Now, $\vert z_n \vert \gt 0$ and $\vert z_n \vert \in \mathbb {R}, \epsilon/(\vert z_n \vert) \gt 0$

$\Rightarrow \vert 1/z_n - 1/z \vert \lt \epsilon $ for all $n\in\mathbb{N}$ such that $\epsilon \gt 0 $

Thus, $(1/z_n) \to (1/z)$

Q. E. D.

However, my fellow- students found it 'quite incomprehendible and jumbled'.

MY question is -

1. Is my proof correct ? If not, why ?
2. If yes, how do I make it clearer ?

Edit-

The earlier edit by zzussee had somewhat changed my intent regarding the last part of the proof.

I had actually intended - $\vert 1/z_n - 1/z \vert \to \upsilon $ , where $\upsilon := { \epsilon \over (\vert z_n \vert )} $,

where $\upsilon $ should be position as both $\epsilon $ and $ z_n $ are positive.

Of course, this does not invalidate the answer by @Jonas Lenz but it explains both his and @ Lucas' comments about $\epsilon $ in the last part of the proof.

Answer 1

From my point of view it is not correct but it contains the correct ideas.

It would proceed as follows. Every proof concerning convergence should start with the obligatory: Let $\varepsilon>0$. Our aim is find some $N \in \mathbb{N}$ such that $\left\vert \frac{1}{z_n}-\frac{1}{z}\right\vert<\varepsilon$ for all $n\geq N$.

As $(z_n)$ converges to $z$ for each $\varepsilon'$ there is $N_1\in \mathbb{N}$ such that $|z_n-z|<\varepsilon$ for $n\geq N_1$. Hence (as showed correctly), we have \begin{align*} \left\vert \frac{1}{z_n}-\frac{1}{z}\right\vert < \frac{\varepsilon'}{|z_n|\cdot |z|} \end{align*} but not for all $n\in \mathbb{N}$ but only for $n\geq N_1$ (I used a different epsilon here as this will need to be chosen correctly in order to show the desired estimate.)

Now, this is almost good but the $n$-dependence in the denominator is not so nice. But as $(z_n)$ converges to $z$ we can find $N_2\in \mathbb{N}$ such that $|z_n|>\frac{1}{2}|z|>0$ for all $n\geq N_2$.

Plugging this into our previous estimate and choosing $\varepsilon':=\varepsilon \cdot \frac{1}{2}|z|^2$ we obtain

\begin{align*} \left\vert \frac{1}{z_n}-\frac{1}{z}\right\vert <\frac{\varepsilon \cdot 2|z|^2}{2\vert z\vert ^2}=\varepsilon \end{align*}

for all $n\geq N:=\max(N_1,N_2)$, which completes the proof.

I actually do not understand the last part of your proof just before you conclude the convergence.