$\displaystyle {prove \ that \ if \ \mid x_{n+1} - x_n \mid \leq \frac{1}{3^n} , \forall n \in N ,\ then \ x_n \ is \ a \ cauchy \ sequence\ .}$

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- Sep 7th 2009, 11:23 AMflower3real analysis
$\displaystyle {prove \ that \ if \ \mid x_{n+1} - x_n \mid \leq \frac{1}{3^n} , \forall n \in N ,\ then \ x_n \ is \ a \ cauchy \ sequence\ .}$

- Sep 7th 2009, 12:05 PMPlato
If $\displaystyle J>K$ then $\displaystyle \left| {x_J - x_K } \right| \leqslant \sum\limits_{k = K}^{J - 1} {\left| {x_{k + 1} - x_k } \right|} \leqslant \sum\limits_{k = K}^{J - 1} {\frac{1}{{3^k }}} $

- Sep 7th 2009, 02:18 PMJG89
Just as you said, $\displaystyle |x_J - x_K| \le \sum_{k = K}^{J-1} \frac{1}{3^k} $.

That's a finite sum, and so for large enough k, the entire sum should go to 0. But if we're trying to prove that $\displaystyle |x_J - x_K| < \epsilon $ for**all**J, K > k, then isn't it possible to hold K fixed and take J so large that the sum doesn't actually approach 0?

Let me know if my question isn't clear enough. - Sep 7th 2009, 02:36 PMPlato
- Sep 7th 2009, 02:45 PMJG89
But the sum converges to 3/2. So $\displaystyle |x_J - x_K| \le 3/2 $.

Aren't we trying to show that |x_J - x_K| goes to 0? - Sep 7th 2009, 03:00 PMPlato
No, you have it all wrong.

$\displaystyle \sum\limits_{k = K}^{J} {\frac{1}{3^k}} $ is the difference in the partial sums, $\displaystyle S_J~\&~S_{K+1}$.

Because $\displaystyle \left( {S_n } \right)$ is a Cauchy sequence that difference can be made as small as necessary. - Sep 7th 2009, 05:16 PMJG89
Wow, I had worked that out but didn't realize I was working out the partial sums...no excuses though, that was a silly mistake. Thanks for the help Plato