I'm doing exercises. In the related book, there is a claim. Is this right? I'm not sure.
For a sequence $\{a_n\}$, there exists a limit $a$ such that $\lim_{n\to\infty} a_n=a$ if and only if for any $p\in \Bbb N$, $\lim_{n\to\infty} |a_{n+p}-a_n|=0$
If not, could you kindly give some counterexamples?
Thanks.
$a_n = \displaystyle\sum_{k=1}^{n}\frac{1}{k}$ diverges, and $\displaystyle\lim_{n\to \infty} \sum_{k=n+1}^{n+p}\frac{1}{k}=0$.
The question Is there a divergent sequence such that $(x_{k+1}-x_k)\rightarrow 0$? is very similar. It asks for a sequence $(a_n)$ that has no limit but such that $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$. If $(a_n)$ is any such sequence, then it is also a counterexample for your question, because $$|a_{n+p}-a_n|\leq |a_{n+p}-a_{n+p-1}|+|a_{n+p-1}-a_{n+p-2}|+\cdots+|a_{n+1}-a_n|,$$ and the last expression goes to $0$ because it is a sum of $p$ things that all go to $0$ as $n$ goes to $\infty$.
We give an example where the limit, even in the "extended" ($\infty$) sense does not exist: $$0, 1, \frac{1}{2}, 0, \frac{1}{3},\frac{2}{3}, 1, \frac{3}{4}, \frac{2}{4},\frac{1}{4},0, \frac{1}{5},\frac{2}{5},\frac{3}{5}, \frac{4}{5}, 1, \frac{5}{6},\frac{4}{6}, \frac{3}{6}, \frac{2}{6},\frac{1}{6},0, \frac{1}{7},\frac{2}{7}\dots.$$ Every real number between $0$ and $1$ is an accumulation point of the above sequence.
