Problem with showing that every Contractive Sequence is Cauchy

Question

Here's the proof which shows that every contractive sequence satisfies the Cauchy criterion, and I don't seem to be convinced with it.

My doubts:

1. Why is the fact that the RHS tends to zero on approaching larger $n$, sufficient to say that we can always find a natural number (say $K$) such that the Cauchy criterion is met for $m,n \ge K$
2. What exactly is the epsilon we choose, here? Is it the expression on the RHS? If so, we should be able to show that the RHS ranges over all real numbers, because we're concerned with all $\varepsilon >0$

Answer 1

Instead of explicitly demonstrating some natural $K$ making the Cauchy condition satisfied, the proof is indirectly doing that by using a known fact from Real Analysis that the sequence $\{C^{n - 1}\}_{n \in \Bbb Z_+}$ tends to $0$ as $\Bbb Z_+ \ni n \to \infty$ as long as $0 < C < 1$ (which indeed holds for contractive sequences).

In other words, the natural number $K$ is "secretly" being chosen like this:

If $x_2 = x_1$, then there is nothing to prove. So assume $|x_2 - x_1| > 0$.

Now fix some $\varepsilon > 0$. And note that because $C^{n - 1} \to 0$ as long as $0 < C < 1$, there must be some natural $K$ such that $\Bbb Z_+ \ni \forall n > K$ we have $$|C^{n - 1}| = C^{n - 1} < \varepsilon \frac{1 - C}{|x_2 - x_1|}$$ (note $1 - C > 0$ as $0 < C < 1$).

Hence continuing the proof from the last inequality, we get that for our epsilon above and for all naturals $m > n > K$: $$ |x_m - x_n| < C^{n - 1}\frac{1}{1 - C}|x_2- x_1| < \varepsilon \frac{1-C}{|x_2 - x_1|} \cdot \frac{1}{1 - C}|x_2 - x_1| = \varepsilon $$ demonstrating that the sequence is Cauchy.

Answer 2

Take $\varepsilon>0$. Take $K\in\mathbb N$ such that$$n\geqslant K\implies C^{n-1}\left(\frac1{1-C}\right)\lvert x_2-x_1\rvert<\varepsilon.$$Then, if $m,n\geqslant K$, you have $3$ possibilities:

1. $m=n$: then $\lvert x_m-x_n\rvert=0<\varepsilon$.
2. $m>n$: then $\lvert x_m-x_n\rvert\leqslant C^{n-1}\left(\frac1{1-C}\right)\lvert x_2-x_1\rvert<\varepsilon$.
3. $m<n$: then $\lvert x_m-x_n\rvert\leqslant C^{m-1}\left(\frac1{1-C}\right)\lvert x_2-x_1\rvert<\varepsilon$.

Answer 3

Take the sequence $$h_n=C^{n-1}\left(\frac{1}{1-C}\right)|x_2-x_1|$$ Now, $h_n\rightarrow 0$ as $n\rightarrow\infty$, which in other sense, means that $h_n$ converges to $0$.

Thus, $\forall\epsilon>0,$ $\exists$ $k\in\mathbb{N}$ such that $|h_n-0|<\epsilon$ $\forall n\geq k$.... $(i)$

So combining the result that $|x_m-x_n|\leq h_n \forall n\in\mathbb{N}$ and $(i)$, we get,

$\forall\epsilon>0, \exists k\in\mathbb{N}$ such that $|x_m-x_n|\leq|h_n|<\epsilon$ $\forall n\geq k$$ $$\implies \forall\epsilon>0, \exists k\in\mathbb{N}$ such that $|x_m-x_n|<\epsilon$ $\forall m,n\geq k$ which proves that this sequence is a Cauchy sequence.