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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)?

Thanks in advance!

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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.

Concerning your question, you proved that the difference between two elements of your sequence is bounded in absolute value by the difference of two elements of another sequence, which you know is converging.

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  • $\begingroup$ that makes sense! thank you :) $\endgroup$ – clovis Oct 15 '18 at 20:02

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