Existence of non-convergent sequence with first differences converging to zero

Question

Does there exist a sequence $x_n$ such that $|x_n - x_{n+1}| \rightarrow 0$ but $x_n$ isn't convergent?

I was looking for such a sequence, but I can't find one. Maybe if $|x_n - x_{n+1}| \rightarrow 0$, then $x_n$ is Cauchy sequence?

Answer 1

Consider the sequence $$\left\{1,~2,~2\frac{1}{2},~3,~3\frac13,~3\frac23,~4,~4\frac14,~4\frac24,~4\frac34,~\ldots\right\}$$

which has $\left|x_n-x_{n+1}\right|\to0$ but $x_n\to\infty\notin\mathbb{R}$.

For a sequence to be Cauchy it must be the case that for any $\varepsilon>0$, there must exist a $T$ such that for any $t_1,t_2>T$, $x_{t_2}\in B_\varepsilon\left(x_{t_1}\right)$, where $B_r\left(u\right)$ denotes the ball of radius $r$ about $u$. As $\mathbb{R}^N$ is a complete space, any Cauchy sequence is convergent.

Answer 2

Let $x_n = \sum_{k=1}^n 1/k$. Then $|x_n-x_{n+1}|=1/(n+1)\to0$ as $n\to\infty$, but $x_n\to\infty$.

Answer 3

For $$x_n = \sum_{i=1}^n \frac{1}{i} $$ we know that $x_n$ diverges. But $$x_{n+1}-x_n =\sum_{i=1}^{n+1} \frac{1}{i}-\sum_{i=1}^n \frac{1}{i} =\frac{1}{n+1}$$ which converges.