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.
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: Consider $a_n= 1 + \frac{1}{2} + \ldots + \frac{1}{n}$
Then $\{a_n\}$ is divergent, but for $p > 0$ we have $$a_{n+p} - a_n = \frac{1}{n+1} + \ldots + \frac{1}{n+p} \leq \frac{p}{n+1} \to 0$$
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 $\lvert x_{n+1}-x_n\rvert = \ln(1+\frac{1}{n}) \xrightarrow[n\to\infty]{} 0$.
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.