real analysis

• September 7th 2009, 11:23 AM
flower3
real analysis
${prove \ that \ if \ \mid x_{n+1} - x_n \mid \leq \frac{1}{3^n} , \forall n \in N ,\ then \ x_n \ is \ a \ cauchy \ sequence\ .}$
• September 7th 2009, 12:05 PM
Plato
If $J>K$ then $\left| {x_J - x_K } \right| \leqslant \sum\limits_{k = K}^{J - 1} {\left| {x_{k + 1} - x_k } \right|} \leqslant \sum\limits_{k = K}^{J - 1} {\frac{1}{{3^k }}}$
• September 7th 2009, 02:18 PM
JG89
Just as you said, $|x_J - x_K| \le \sum_{k = K}^{J-1} \frac{1}{3^k}$.

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 $|x_J - x_K| < \epsilon$ 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.
• September 7th 2009, 02:36 PM
Plato
Quote:

Originally Posted by JG89
Just as you said, $|x_J - x_K| \le \sum_{k = K}^{J-1} \frac{1}{3^k}$.
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 $|x_J - x_K| < \epsilon$ 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?

But the series $\sum_{k = 1}^{\infty} \frac{1}{3^k}$ converges.
Therefore the sequence of partial sums converges; so it is a Cauchy sequence.
That means we done.
• September 7th 2009, 02:45 PM
JG89
But the sum converges to 3/2. So $|x_J - x_K| \le 3/2$.

Aren't we trying to show that |x_J - x_K| goes to 0?
• September 7th 2009, 03:00 PM
Plato
Quote:

Originally Posted by JG89
But the sum converges to 3/2. So $|x_J - x_K| \le 3/2$.
Aren't we trying to show that |x_J - x_K| goes to 0?

No, you have it all wrong.
$\sum\limits_{k = K}^{J} {\frac{1}{3^k}}$ is the difference in the partial sums, $S_J~\&~S_{K+1}$.
Because $\left( {S_n } \right)$ is a Cauchy sequence that difference can be made as small as necessary.
• September 7th 2009, 05:16 PM
JG89
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