# Example of a sequence that is not Cauchy

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.

• 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. Oct 12, 2015 at 20:37
• The sequence $x_k=\log k$ is the standard example. Oct 12, 2015 at 20:42
• how does $\log(k+1)-\log k\to 0$? @Crostul Oct 12, 2015 at 20:44
• @JennieDurham $\ln\frac{k+1}{k} = \ln(1+\frac{1}{k}) \sim \frac{1}{k}$. Oct 12, 2015 at 20:46

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.

Hint: Consider $a_n= 1 + \frac{1}{2} + \ldots + \frac{1}{n}$

Then $\{a_n\}$ is divergent, but for $p > 0$ we have $$a_{n+p} - a_n = \frac{1}{n+1} + \ldots + \frac{1}{n+p} \leq \frac{p}{n+1} \to 0$$

• You proved something stronger than the required condition, by the way. You only needed to show that $|a_{k+1}-a_k|$ goes to $0$. Oct 12, 2015 at 20:50
• Take $p =1$ and it's done. Oct 12, 2015 at 20:50

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 $\lvert x_{n+1}-x_n\rvert = \ln(1+\frac{1}{n}) \xrightarrow[n\to\infty]{} 0$.

• 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! Jan 11, 2020 at 22:31