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I want an example of a sequence that satisfies $\mid x(n) - x(n-1)\mid \to 0$ but not Cauchy ? I tried to find such sequence $x(n)=1/2,1/3,1/2,1/3,1/4,1/2,1/3,1/4,1/5,,,,$ it's not Cauchy since it is not converges (has infinitely limit points) and the difference between any successive terms goes to zero .. I asked my doctor but he said the difference is constant (not zero)since the sub-sequences here are constants So can any one gives me a hint for this problem :)? thanks in advance ^^

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Try $$ x_n =\sum_{k=1}^n \frac{1}{k}. $$

A slightly different example would be $x_n =\ln n$.

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  • $\begingroup$ Thanks for replying :)),,,yes that's absolutely true the difference between any successive terms is 1/n which converges to zero while the infinte series is divergent,,Thanks again:) $\endgroup$ – Sara Suradi Oct 24 '15 at 7:45
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Another example is $$x_n = \sin \sqrt n \, .$$ As in Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$., the set of limit points is the entire interval $[-1, 1]$, and using the mean value theorem it is easy to see that $$ | x_{n} - x_{n-1} | \le \frac {1}{2 \sqrt {n-1}} \to 0 \, . $$

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$1, 1/2,0,1/3,2/3,1,3/4,2/4,1/4, 0, \dots.$

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