Let $r_n$ be a sequence of all rational numbers from $(0,1)$. Show that the series $\sum_{n=2}^\infty|r_n-r_{n-1}|$ diverges Let $r_n$ be a sequence of all rational numbers in $(0,1)$. Show that the sum $\sum_{n=2}^\infty|r_n-r_{n-1}|$ diverges.
I've seen this problem quite a few times and to be honest I really don't know how to deal with it. I would really appreciate some help so I can understand this.
Does it have anything to do with the absolute convergence?
 A: Hint. Note that if $\sum_{n=2}^\infty|r_n-r_{n-1}|$ is convergent then for $\epsilon>0$ there is $R>0$ such that  for $M\geq N>R$,
$$|r_M-r_N|\leq \sum_{n=N+1}^{M}|r_n-r_{n-1}|<\epsilon$$
which means that $\{r_n\}_{n\geq 1}$ is a Cauchy sequence and therefore convergent. Show that this contradicts the fact that $\{r_n\}_{n\geq 1}$ is dense in $(0,1)$.
A: You may also use a partitioning argument. Let $m\geq 3$ and let us split $(0,1)$ as $I_1\cup I_2\cup\ldots\cup I_m$, where $I_1=\left(0,\frac{1}{m}\right), I_2=\left[\frac{1}{m},\frac{2}{m}\right),\ldots,I_m=\left[\frac{m-1}{m},1\right)$. There is some $N=N(m)$ such that the elements of $\{r_1,\ldots,r_N\}$ occupy all the subintervals. We may assume that $r_a\in I_1$ and $r_b\in I_m$ with $a<b\leq N$. By the triangle inequality
$$ |r_{a+1}-r_a|+|r_{a+2}-r_{a+1}|+\ldots+|r_{b}-r_{b-1}| \geq r_b-r_a \geq \frac{1}{m}.$$
Now we may remove the elements $r_1,\ldots,r_M$ from the sequence, perform a re-indexing and increase $m$ by one. It follows that
$$ \sum_{n\geq 1}|r_{n+1}-r_n|\geq \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots = +\infty.$$
In a geometric fashion: by the density of $\mathbb{Q}$ in itself we have that $\{\ldots,\frac{1}{8},\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{7}{8},\ldots\}$ are accumulation points for the $\{r_n\}_{n\geq 1}$ sequence, hence by assuming that $\sum_{n\geq 1}|r_{n+1}-r_n|$ is convergent and by invoking the triangle inequality we have that the previous series is greater of equal than $\frac{1}{\pi}$ times the length of the following pseudo-spiral

which clearly has an unbouded length.
A: Assume your series converges, then the series $\sum\limits_{n=2}^\infty r_n-r_{n-1}$ converges. But $\sum\limits_{n=2}^N r_n-r_{n-1}=r_N-r_1$, so that would imply the convergence of the sequence $\{r_n\}$. However, by definition of the real numbers, every number in $[0,1]$ is a limit point of some sequence of rational numbers, so $\{r_n\}$ must have infinitely many limit points, so it can't converge. Thus we have a contradiction.
A: $$\sum_{n=2}^\infty\left|r_n-r_{n-1}\right|>\sum_{n=3}^\infty\left|\dfrac{2}{n}-\dfrac{1}{n}\right|=\infty$$
