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Thread: cauchy sequence

  1. #1
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    cauchy sequence

    Show directly that the following sequence is not cauchy :
    $\displaystyle x_n=ln \ n$
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    Senior Member Shanks's Avatar
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    Hint:what is $\displaystyle \left|x_{2n}-x_{n}\right|$?
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    Quote Originally Posted by flower3 View Post
    Show directly that the following sequence is not cauchy :
    $\displaystyle x_n=ln \ n$
    It is easy to show that if $\displaystyle N\in\mathbb{Z}^+~\&~k\in\mathbb{Z}^+$ then $\displaystyle \frac{k}{N+k}\le \ln(N+k)-\ln(N)$.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    Show directly that the following sequence is not cauchy :
    $\displaystyle x_n=ln \ n$
    Quote Originally Posted by Shanks View Post
    Hint:what is $\displaystyle \left|x_{2n}-x_{n}\right|$?
    I think what Shanks meant was consider that for every $\displaystyle n>1$ it is obviously true that $\displaystyle n^2>n$, but $\displaystyle \left|\ln\left(n^2\right)-\ln(n)\right|=\left|2\ln(n)-\ln(n)\right|=\left|\ln(n)\right|=\ln(n)$, which is clearly greater than any $\displaystyle \varepsilon>0$ for sufficiently large $\displaystyle n$.
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    Quote Originally Posted by flower3 View Post
    Show directly that the following sequence is not cauchy :
    $\displaystyle x_n=ln \ n$
    cancel this post
    Last edited by xalk; Dec 20th 2009 at 09:48 AM.
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