Show directly that the following sequence is not cauchy :

$\displaystyle x_n=ln \ n$

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- Dec 19th 2009, 08:52 AMflower3cauchy sequence
Show directly that the following sequence is not cauchy :

$\displaystyle x_n=ln \ n$ - Dec 19th 2009, 09:32 AMShanks
Hint:what is $\displaystyle \left|x_{2n}-x_{n}\right|$?

- Dec 19th 2009, 09:32 AMPlato
- Dec 19th 2009, 09:44 PMDrexel28
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 $\displaystyle n$. - Dec 20th 2009, 09:37 AMxalk