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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?

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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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