Bounded sequence which is not convergent, but differences of consecutive terms converge to zero

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.

• 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 $|x_n-x_m|<\varepsilon$. Sep 16 '15 at 2:46
• Sep 16 '15 at 6:23

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.
• So showing $\neq$ proving?
• 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. Sep 16 '15 at 15:03
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.