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Give an example of a sequence $(x_n)$ of real numbers, where $\displaystyle\lim_{n\to+\infty}|x_n-x_{n+1}|=0$, but $(x_n)$ is not a Cauchy sequence

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Try $x_n=\sqrt n$. – Davide Giraudo Oct 25 '11 at 19:42
Alternatively, consider $x_n$ = log($n$). – Grasshopper Oct 25 '11 at 19:47
This is not the simplest answer, but you can also $x_n$ to be the $n^{th}$ partial sum of a divergent series whose terms approach $0$. (You can even get examples where the terms are nonnegative and monotonically decreasing.) For e.g., the harmonic series fits the bill. – Srivatsan Oct 25 '11 at 19:49
Or take $x_n$ to be the $n$th partial sum of any divergent series $\sum a_k$ such that $a_k\to0$ as $k\to\infty$. – Nick Strehlke Oct 25 '11 at 19:50
@SrivatsanNarayanan I see you beat me by 35 seconds! :) – Nick Strehlke Oct 25 '11 at 19:51
up vote 7 down vote accepted

Posting my comment as an answer.

You can take $x_n$ to be the $n^{th}$ partial sum of any divergent series $\sum_{k} a_k$ such that $a_k \to 0$ as $k \to \infty$. In fact, we can even produce such an example with the additional restriction that the sequence $(a_k)$ is positive and monontonically decreasing. The canonical example of such a series is the harmonic series $\sum_k \frac{1}{k}$.

Actually, the above description is complete in the sense that if $x_n$ is any sequence satisfying the OP's requirements, then the series $\sum_k a_k$ defined by $a_n = x_n - x_{n-1}$ is such that $a_k \to 0$ as $k \to \infty$ and yet the series is divergent.

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One problem with your solution - it is very possible that "divergent series" is a concept not familiar to someone asking a homework question about Cacuhy sequences... – Gadi A Oct 25 '11 at 20:45
@Gadi Thanks for pointing it out. Yes, I too find the solution unsatisfactory, and that's one reason for posting it as CW. But I just felt that the connection is interesting and deserves to be stated somewhere (for other readers, if not for the OP). Also I find the description in terms of series is more intuitive than the condition $x_n - x_{n-1} \to 0$ as $n \to \infty$ (even though these two say the same thing). – Srivatsan Oct 25 '11 at 20:50

Take $x_n=\sqrt n$. Since $|x_{n+1}-x_n|=\left|\sqrt{n+1}-\sqrt n\right|=\left|\frac {n+1-1}{\sqrt{n+1}+\sqrt n}\right|\leq \frac 1{\sqrt n}$, we have $\displaystyle\lim_{n\to+\infty}x_{n+1}-x_n=0$. But we have $|x_{2n}-x_n|=\sqrt 2\sqrt n-\sqrt n=(\sqrt 2-1)\sqrt n$, hence $(x_n)_n$ cannot be a Cauchy sequence.

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$x_n$ is unbounded so it cannot be a Cauchy sequence, thus you don't have to do these computations in the second sentence. – Damian Sobota Oct 25 '11 at 21:17
I agree, but the computations to show this statement are not more complicated than mine. – Davide Giraudo Oct 25 '11 at 21:20

I would like to add that one can construct a bounded sequence $\{x_n\}$ satisfying $| x_n - x_{n+1}|\rightarrow 0$ that is not Cauchy. E.g., take: $$ 0, {1\over2}, 1, {2\over3}, {1\over3}, 0 ,{1\over4}, {2\over4}, {3\over4}, 1, {4\over5}, {3\over5}, {2\over5}, {1\over 5}, 0,{1\over6}, \ldots $$

Please forgive me if I've ``necroed'' a thread; I'm new here...

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