Prove that if $\sum|a_n-a_{n+1}| < \infty$, then the sequence $(a_n)$ converges, but the converse is not true.

I thought I had the first part, but just realized I made a mistake: I said that we have $\sum|a_n-a_{n+1}|$ converges by the Monotone Convergence Theorem, so the sequence $(|a_n-a_{n+1}|)$ converges to $0$, and this means that $(a_n)$ is a Cauchy sequence which implies $(a_n)$ converges. I realized that from what I have, I can't conclude that $(a_n)$ is a Cauchy sequence since I only know about the absolute difference between $a_n$ and $a_{n+1}$, and not any arbitrary $a_n$ and $a_m$, where $m,n >$ some $N\in\mathbb{N}$. I also tried the other direction, hoping to find some alternating $(a_n)$sequence that converged, but for which $(|a_n-a_{n+1}|)$ did not converge to $0$ (so the series would have to diverge), but had no luck. So now I'm not sure how to proceed in either direction. Any hints?


For the forward direction, use the triangle inequality and the fact that your sum converges to show $(a_n)$ is Cauchy.

Towards this end, note for any $n$, $k$ that $$a_n-a_{n+k}=(a_n-a_{n+1})+(a_{n+1} -a_{n+2})+ \cdots+(a_{n+k-2} -a_{n+k-1})+(a_{n+k-1} -a_k).$$

For the converse, consider the sequence $(1,0, 1/2,0,1/3,0,\ldots)$.


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