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I try to prove this by proving $(a_n)$ is a Cauchy sequence. First, since $s_n = \sum|a_n - a_{n+1}| < \infty$, $s_n$ converges and thus $$\lim_{n\rightarrow \infty} |a_n - a_{n+1}| = 0$$ Then, for every $\epsilon > 0$, let $n_0 \in \mathbb{N}$ such that $\forall n \geq n_0, |a_n - a_{n+1}| < \frac{\epsilon}{n_0}$. Then I have $$|a_n - a_m| \leq |a_n - a_{n+1}| + \cdots+|a_m-1 + a_m| \\ < (m-n-1)\frac{\epsilon}{n_0}$$ Here I got stuck, what if $\frac{m-n-1}{n_0}>1$? How can I rewrite this equation so that it will be exactly less than $\epsilon$?Can anyone give me some suggestions?

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    $\begingroup$ $\lim_{n\rightarrow \infty} |a_n - a_{n+1}| = 0$ is not enough to imply convergence, so your approach cannot work. Consider $a_n = \log(n)$, so $|a_{n+1} -a_n| \to 0$ $\endgroup$
    – leonbloy
    Sep 9, 2015 at 2:39

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Observe that $$a_n=a_1+\sum_{k=2}^n(a_k-a_{k-1}).$$ Since the series in question converges absolutely, it converges. So the right hand side has a limit and hence the left hand side must as well.

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