A sequence of real numbers such that $\lim_{n\to+\infty}|x_n-x_{n+1}|=0$ but it is not Cauchy

Solution 1:

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: $$ 0, {1\over2}, 1, {2\over3}, {1\over3}, 0 ,{1\over4}, {2\over4}, {3\over4}, 1, {4\over5}, {3\over5}, {2\over5}, {1\over 5}, 0,{1\over6}, \ldots $$

Solution 2:

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.

Solution 3:

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.