5
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Counterexample $a_n=\sum_{k=1}^{n}\frac{1}{k}$ $\endgroup$ – Angela Richardson Jun 4 '13 at 6:33
  • $\begingroup$ This is an application of the $\epsilon, \delta$ formulation of the limit. $\endgroup$ – ThisIsNotAnId Jun 4 '13 at 6:53
12
$\begingroup$

$a_n = \displaystyle\sum_{k=1}^{n}\frac{1}{k}$ diverges, and $\displaystyle\lim_{n\to \infty} \sum_{k=n+1}^{n+p}\frac{1}{k}=0$.

$\endgroup$
3
$\begingroup$

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 $\infty$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.