Question:
Find an example such that $|x_{k+1}-x_k|\to 0$, but the sequence $x_k$is not Cauchy.
Any help would be appreciated.
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asked Oct 12, 2015 at 20:33
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3$\begingroup$ Hint: What if $x_k$ represented the $k$th partial sum of a divergent series? Then $x_{k+1} - x_k$ would be the $(k+1)$th term of the sequence, so we are left to finding a convergent sequence whose partial sums diverge. $\endgroup$– user217285Oct 12, 2015 at 20:37
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1$\begingroup$ The sequence $x_k=\log k$ is the standard example. $\endgroup$– CrostulOct 12, 2015 at 20:42
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$\begingroup$ how does $\log(k+1)-\log k\to 0$? @Crostul $\endgroup$– Jennie DurhamOct 12, 2015 at 20:44
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1$\begingroup$ @JennieDurham $\ln\frac{k+1}{k} = \ln(1+\frac{1}{k}) \sim \frac{1}{k}$. $\endgroup$– Clement C.Oct 12, 2015 at 20:46
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3 Answers
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The sequence of partial sums $(H_n)_{n\geqslant 1}$ of the harmonic series defined by $$ H_n = \sum_{k=1}^n \frac{1}{k} $$ verifies $H_{n+1} - H_n = 1/(n+1) \xrightarrow[n\to+\infty]{} 0$, but $$ H_{2n} - H_n = \frac{1}{n+1} + \dots + \frac{1}{2n} \geqslant \frac{n}{2n} = \frac{1}{2} $$ so $(H_n)$ is not a cauchy sequence.
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Hint: Consider $a_n= 1 + \frac{1}{2} + \ldots + \frac{1}{n}$
Then $\{a_n\}$ is divergent, but for $p > 0$ we have $$a_{n+p} - a_n = \frac{1}{n+1} + \ldots + \frac{1}{n+p} \leq \frac{p}{n+1} \to 0$$
answered Oct 12, 2015 at 20:41
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$\begingroup$ You proved something stronger than the required condition, by the way. You only needed to show that $|a_{k+1}-a_k|$ goes to $0$. $\endgroup$ Oct 12, 2015 at 20:50
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Hint: any sequence of that sort cannot be convergent, since any convergent sequence is Cauchy. Intuitively, any sequence diverging "slowly enough" to e.g. $\infty$ would thus do the trick.
Take for instance $x_n = \ln n$. $(x_n)$ is not Cauchy (it diverges; yet $\lvert x_{n+1}-x_n\rvert = \ln(1+\frac{1}{n}) \xrightarrow[n\to\infty]{} 0$.
answered Oct 12, 2015 at 20:43
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$\begingroup$ Thanks, this example is far better than $H_n$, since the computation of $x_{2n} - x_n = \ln(2n) - \ln (n) = ln(2) + ln(n) - ln(2) = ln(2)$ provides an equality rather than an inequality. Plus it's easier to note that it diverges! $\endgroup$ Jan 11, 2020 at 22:31