9
$\begingroup$

I have a question that says "Show that there is a bounded sequence $x_n$ which is not convergent but has the property that $x_n - x_{n+1} \to 0$ as $n \to 0$.

What does this mean? Do I need to come up with an example or does the problem actually want me to prove such proposition?

By the way, I see that this sequence looks like Cauchy because of $x_n - x_{n+1} \to 0$ as $n \to 0$, but it is obviously not.

$\endgroup$
2
  • 1
    $\begingroup$ Here you are taking consecutive elements in a sequence; the Cauchy property says that for any $\varepsilon>0$, there exists an $N$ so that for all $m,n>N$ we have $|x_n-x_m|<\varepsilon$. $\endgroup$
    – Clayton
    Sep 16 '15 at 2:46
  • $\begingroup$ See also math.stackexchange.com/questions/359344/… $\endgroup$ Sep 16 '15 at 6:23
16
$\begingroup$

A single example will do the job. Try this: $$0,1,1/2,0,1/3,2/3,1,3/4,2/4,1/4,0,1/5,2/5,3/5,4/5,1,5/6,4/6,3/6,2/6,1/6,0,1/7,\dots.$$

We travel back and forth between $0$ and $1$, first with a giant step, then with two steps of length $1/2$, then with $3$ steps of length $1/3$, and so on. Note that every real number between $0$ and $1$ is a limit of a subsequence of our sequence.

$\endgroup$
4
  • $\begingroup$ So showing $\neq$ proving? $\endgroup$
    – user247618
    Sep 16 '15 at 2:51
  • 3
    $\begingroup$ This shows that there is a sequence with the properties asked for in the question. Usually the simplest way to show that there is an object with specified properties is to exhibit such an object. (Sometimes one gives an indirect proof of existence.) $\endgroup$ Sep 16 '15 at 2:59
  • $\begingroup$ In this case (and usually in mathematics), showing = proving. One way (usually the simplest way) to prove that something exists is to exhibit one example. $\endgroup$ Sep 16 '15 at 14:36
  • 1
    $\begingroup$ About your Cauchy sequence remark: For Cauchy sequence, we need to show that for any $\epsilon$, there is an $N$ such that $|a_{m}-a_n|\lt \epsilon$ whenever $m$ and $n$ are greater than $N$ . This question, among other things, shows that the condition there is an $N$ such that $|a_{n+1}-a_n|\lt \epsilon$ whenever $n\gt N$ is not enough to ensure the sequence is Cauchy. $\endgroup$ Sep 16 '15 at 15:03
11
$\begingroup$

They're asking for a sequence where successive terms get closer to each other but don't converge on a final value. An example that comes to mind would be $\sin(\sqrt{n})$. The gap between successive values of $\sqrt{n}$ gets smaller, but obviously the series is going to oscillate.

$\endgroup$

Your Answer

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