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I want to produce a divergent sequence for which $|x_n - x_{n-1}| \to 0$. So far, I've only been able to show that $$\frac{x_n}{n} \to 0$$, which doesn't really help.

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    $\begingroup$ Here is a related question. $\endgroup$ – David Mitra Sep 19 '14 at 15:54
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    $\begingroup$ $\ln1=0, \ln2, \ln3, \ln4,$ etc. $\endgroup$ – Akiva Weinberger Sep 19 '14 at 16:25
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    $\begingroup$ @columbus8myhw Answers should be posted as answers, not comments. $\endgroup$ – David Richerby Sep 20 '14 at 11:59
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Use the sequence $(1,1+\frac 12,1+\frac 12+\frac 13,\ldots )$.

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    $\begingroup$ I am really considering hitting myself for not thinking of this. Many thanks. $\endgroup$ – user41281 Sep 19 '14 at 15:48
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Another example would be $x_n=\ln n$, since

$\ln n\rightarrow\infty$ but $\ln n-\ln(n-1)=\ln\frac{n}{n-1}\rightarrow\ln 1=0$.

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In general, let $\{x_n\}$ be a divergent sequence with $\lim_{n\rightarrow\infty}x_n/n=0$; if $\lim_{n\rightarrow\infty} x_n-x_{n-1}=L,$ consider $\{y_n:=x_n-nL\}.$ If $\{y_n\}$ converges, then so does $$\{y_n/n=y_n\cdot \frac{1}{n}=x_n/n-L\},$$ whose limit is therefore, according to a basic theorem in calculus, $$=\lim_{n\rightarrow\infty}y_n\cdot\lim_{n\rightarrow\infty}1/n=0,$$ a contradiction unless $L=0.$
Thus, when $L\not=0,$ the sequence $\{y_n\}$ diverges, with $$\lim_{n\rightarrow\infty} y_n-y_{n-1}=\lim_{n\rightarrow\infty} x_n-x_{n-1}-L=0.$$
This shows that what you have tried helps, in fact.
Hope this helps.

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