Using definition of Cauchy sequence in proof of Banach Fixed point Theorem I'm going through the proof of Banach Fixed Point for a metric space $(X,d)$ in this pdf, the step I having trouble with in is the one after the geometric series is summed:
$$ d(x_n , x_m )  < \frac{ \beta^n}{1- \beta} d(x_0,x_1)$$
The sequence $x_n$ is defined as $x_n = Tx_{n-1}$ where $T$ is the contraction mapping. My doubt is how the above sequence is proved to be Cauchy. In the pdf, the following is said:

Theorem $\{x_n \}$ is Cauchy: for $\epsilon>0$ , let $N$ be large enough that $\frac{\beta^N }{1- \beta} d(x_0, x_1) <\epsilon$ , which ensures that $n,m>N \implies d(x_n,x_m) < \epsilon$ .

I can't fit this into the definition I know for Cauchy. I understand that a sequence is Cauchy if for every real number $\epsilon>0$ there exists an $N$ such that any $n,m>N$ makes the $x_m$ th and $x_n$th members epislon close. In equations:

$$\text{sequence is Cauchy} \iff \forall \epsilon>0: \exists M: \forall m,n>M : d(x_m,x_n) < \epsilon$$

In the quoted text for the theorem the $N$ they take is the same as the $N$ which appears in the metric function $d(x_n,x_m)$ but don't we need $n>N$..?
 A: So, I hope you're ok with the fact that for all $n<m$, that
\begin{align}
d(x_n,x_m)<\frac{\beta^n}{1-\beta}d(x_0,x_1)\tag{i}.
\end{align}
Ok, so now we prove the sequence is Cauchy. Suppose you're given an arbitrary $\epsilon>0$. Now, since $0\leq \beta<1$, we have that $\lim\limits_{n\to \infty}\beta^n=0$ (this is a basic fact which I leave to you to prove). Therefore, $\lim\limits_{n\to \infty}\frac{\beta^n}{1-\beta}d(x_0,x_1)=0$ as well. So unwinding the definition of limits here, there exists some $M\in\Bbb{N}$ such that for all $n\geq M$,
\begin{align}
\frac{\beta^n}{1-\beta}d(x_0,x_1)\leq \epsilon\tag{ii}.
\end{align}
Now, let $n,m\geq M$. Without loss of generality (why?), we may as well assume $n<m$. Then,
\begin{align}
d(x_n,x_m)&<\frac{\beta^n}{1-\beta}d(x_0,x_1)\tag{by (i)}\\
&\leq\epsilon.\tag{by (ii)}
\end{align}
So, using the fact $0\leq \beta<1$ (i.e we have a contraction mapping), we showed

For every $\epsilon>0$, there is an $M\in\Bbb{N}$ such that for all $n,m\in\Bbb{N}$, if $n,m\geq M$ then $d(x_n,x_m)< \epsilon$,

which is precisely the condition for a sequence to be Cauchy.
