For a sequence $(a_n)$ of real numbers, $\sum_{n=1}^\infty |a_{n+1}-a_n|$ converges implies $(a_n)_{n\in\mathbb{N}}$ converges.

Question

I know that when the series $\sum_{n=1}^\infty |a_{n+1}-a_n|$ converges, then we have $|a_{n+1}-a_n|\rightarrow 0$
So by using this I was going to prove that the sequence $a_n$ is Cauchy. But couldn't come up with a correct way.

Answer 1

Hint: For every $n \geq 0$, $p \geq 0$, $$|a_{n+p}-a_n| \leq \sum_{k=n}^{\infty}{|a_{k+1}-a_k|}.$$

Answer 2

$S_n:=\sum_{k=1}^{n}|a_{k+1}-a_{k}|.$

Since $S_n$ converges , $S_n$ is Cauchy.

$\epsilon >0$ given, there exists a $n_0$ s.t. for $m \ge n \ge n_0$

$|S_m-S_n| =|\sum_{k=n+1}^{m}|a_{k+1}-a_{k}||$

$=\sum_{k=n+1}^{m}|a_{k+1}-a_k| < \epsilon.$

Show that $(a_n)_{n \in \mathbb{N}}$ is Cauchy.

For $m \ge n \ge (n_0 +1)$ we have:

$|a_m-a_n| \le |a_m-a_{m-1}| +..$

$|a_{m-1}-a_{m-2}|+.......|a_{n+1}-a_n| $

$=S_{m-1}-S_{n-1} \lt \epsilon.$

Hence Cauchy, hence convergent.