Proof that sequence is a Cauchy sequence. Let $s_0$ and $s_1$ be arbitrary real numbers. Suppose $\alpha, \beta >0$ such that $\alpha + \beta = 1$. Define the sequence for $n \geq 1$, $$s_{n+1} = \alpha s_{n} + \beta s_{n-1}. $$ I'm trying to show that the sequence is Cauchy by considering the successive differences, $|s_{n+1} - s_n|$. Clearly, $$|s_{n+1} - s_n| = |\alpha s_n + \beta s_{n-1} - s_n | = \beta |s_n - s_{n-1}|.$$ Since $0 < \beta < 1, |s_{n+1} - s_n| < |s_n - s_{n-1}|$. Therefore as $n \to \infty$,  the difference between subsequent terms decreases.
Does this imply that the sequence is a Cauchy sequence? If so, is there a way of making my argument more rigorous?
 A: Ok, let us finish this with a rigurous $\epsilon$ proof:
You have that $|s_{n+1} - s_n| = \beta |s_n - s_{n-1}|$, so by an easy induction:
$$|s_{n+1} - s_n| = \beta^n |s_1 - s_0|$$
Now let us try to see that the sequence is Cauchy by definition. Let $\epsilon>0$ we want to find some $N\in\mathbb{N}$ such that $|s_n-s_m|<\epsilon$ for all $n,m\geq N$. Without loss of generality we may assume $n\geq m$, so:
$$|s_n - s_m| =|s_n-s_{n-1}+s_{n-1}+\dots-s_{m+1}+s_{m+1}-s_m|\leq\sum_{j=m}^n|s_j-s_{j+1}|=$$
$$=\sum_{j=m}^n\beta^{j-1} |s_1 - s_0|=  |s_1 - s_0|\sum_{j=m}^n\beta^{j-1} $$
Now notice that since $|\beta|<1$ by the geometric series we know that :$$\sum_{j=0}^\infty\beta^{j}<\infty $$
And so:
$$\lim_{r\to\infty}\sum_{j=r}^\infty\beta^{j}=0$$
This implies that $\forall~\epsilon>0$ there exists some $N\in\mathbb{N}$ such that:
$$\sum_{j=n-1}^\infty\beta^{j}<\frac{\epsilon}{|s_1-s_0|}~\forall~n\geq N$$
In particular $\forall~n,m\geq N$ (we assume $n\geq m$) we have:
$$|s_n - s_m| \leq|s_1-s_0|\sum_{j=m}^n\beta^{j-1} \leq |s_1-s_0|\sum_{j=m}^\infty\beta^{j-1}= |s_1-s_0|\sum_{j=m-1}^\infty\beta^{j}<\epsilon$$
Hence, the sequence is Cauchy by definition.
Remark: I am assuming $s_1\neq s_0$, in other case do the same with $s_a$ and $s_{a-1}$ where $a=\inf\lbrace n\geq 1|s_n\neq s_{n-1}\rbrace$.
