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Math Help - real analysis

  1. #1
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    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\ .}
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  2. #2
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    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 }}}
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by JG89 View Post
    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.
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  5. #5
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    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?
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  6. #6
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    Quote Originally Posted by JG89 View Post
    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.
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  7. #7
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    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
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