# Understanding a Cauchy Sequence Proof

I've got some general confusion over Cauchy sequences I'm trying to clear up. Intuitively, I know that the Cauchy theorem lets us prove convergence of a sequence without knowing about the real number L it actually converges to, by establishing that the distance between two points in the sequence is very small, beyond a certain $$N$$. But actually proving something's a Cauchy theorem with the definition isn't as intuitive for me:

I'm trying to prove that the sequence $$(x_n)$$ which satisfies $$|x_{n+1} - x_n| \le r|x_n - x_{n-1}|$$ where $$0 < r <1$$, is a Cauchy sequence.

I started like this:

for $$n > m$$,

$$|x_n-x_m| = r((|x_n - x_{n-1}|+(|x_{n-1}-x_{n-2}|)+...+(|x_{m+1}-x_m|)) =\sum_{i=m+1} r|x_i - x_{i-1}| \lt \sum_{i=m+1} r(r^{i-1}) \lt {\frac{r(r^{m}-r^{n})}{1-r}}$$

but then I'm not sure on how I should proceed or if it's even on the right track (mostly because of that r out front in the numerator).

And I guess at the crux of it all, I'm struggling to understand; what are we actually showing when we apply the Triangle Inequality to these sequences in these proofs (i.e. the right hand side of $$|x_n - x_m|)$$? That the difference of a sequence beyond a certain point is smaller than all of these points in the sequence summed together (thereby helping us prove convergence)?

You are on the right track. You proved that $$|x_n-x_m| \lt {\frac{r(r^{m}-r^{n})}{1-r}}\le {\frac{r^{m+1}}{1-r}} .$$ Now since $$r^m\to 0$$ as $$m\to\infty$$, you have that $${\frac{r^{m+1}}{1-r}}\to 0$$ as $$m\to\infty$$ and so given $$\varepsilon>0$$ you can find $$N$$ such that $${\frac{r^{m+1}}{1-r}}\le \varepsilon$$ for all $$m\ge N$$. Hence, $$|x_n-x_m| \lt \varepsilon$$ for all $$n>m\ge N$$, which shows that you have a Cauchy sequence.