Cauchy sequence proof check

Question

Is this a correct proof of the following problem:

Let $\{ x_n \}$ be a sequence such that there exist $A >0$ and $C \in (0,1)$ for which

$$|x_{n+1}-x_n| \leq AC^n$$

for any $n \geq 1$. Show that $\{x_n\}$ is Cauchy.

My proof (hopefully):

1. The sequence is Cauchy if: $(\forall \epsilon >0)(\exists N)(\forall m,n \geq N)(|x_m -x_n|<\epsilon)$
2. We will be able to select $N$ for any $\epsilon$ selected provided that $AC^n$ goes to zero (i.e. if $AC^n$ goes to zero as $n$ increases then for any selected $\epsilon$ we are able to select large enough $N$, beyond which the absolute difference between $x_m$ and $x_n$ will be epsilon or less).
3. We also need monotnic sequence for the above to be true. In this case the sequence is monotonically decreasing. Because the absolute difference between the subsequent terms in the sequence becomes smaller and smaller ($\lim_{n \rightarrow \infty} C^n = 0$).

Thus, due to $2$ and $3$. The sequence is Cauchy.

Answer 1

$$|x_m-x_n|=|(x_m-x_{m-1})+(x_{n-1}-x_n)+(x_{m-1}-x_{n-1})|\\ |x_m-x_n|\le |x_m-x_{m-1}|+|x_{n-1}-x_n|+|x_{m-1}-x_{n-1}|\\ |x_m-x_n|\le A(C^{m}+C^{n})+|x_{m-1}-x_{n-1}|$$

doing the same idea

$$|x_{m-1}-x_{n-1}|\le A(C^{m-1}+C^{n-1})+|x_{m-2}-x_{n-2}|$$

going on with the same idea we get (supposing that $m\le n$):

$$|x_m-x_n|\le A(C^{m}+C^{m-1}+...+C+C^{n}+C^{n-1}+...+C^{n-m+1})+|x_{0}-x_{n-m}|\quad (1)$$

But

$$|x_k-x_0|=|x_k-x_{k-1}+x_{k-1}-x_0|\le AC^k+|x_{k-1}-x_0|\quad (2)$$

Using $(2)$ in $(1)$ we get:

$$|x_m-x_n|\le A[(C^{m}+C^{m-1}+...+C)+(C^{n}+C^{n-1}+...+C)]=A\left[\frac{C^{m+1}-1}{C-1}+\frac{C^{n+1}-1}{C-1}\right]$$

Once $C^{m+1}-1$ and $C^{n+1}-1$ goes to zero when $m$ and $n$ goes to infinity then we can choose $m$ and $n$ big enough such that

$$\frac{C^{m+1}-1}{C-1}\le\frac{\epsilon}{2A},\\ \frac{C^{n+1}-1}{C-1}\le\frac{\epsilon}{2A}$$

for any $\epsilon$.

So

$$|x_m-x_n|\le \epsilon$$