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Let $a_n$ be a sequence such that $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} |a_n - a_{n+1}| < \infty$. Show that $a_n$ is Cauchy

Attempt:

to show the sequences is Cauchy I have to find $N$ such that for all $m,n \geq N$ one has $|a_m - a_n| < \epsilon$

$$|a_m - a_n| = |a_m - a_{m+1} + a_{m+1} - a_{m+2} + .... - a_{n-1} + a_{n-1} - a_n|$$ $$\leq |a_m - a_{m+1}| + |a_{m+1} - a_{m+2}| + ....+|a_{n-2} - a_{n-1}| + |a_{n-1} - a_n|$$ $$= \sum_{i=m}^{n} |a_i - a_{i+1}|$$

Here is where I am stuck. How can I use the piece I was given to arrive at an Epsilon. Specifically how to use: $\lim_{N\rightarrow \infty} \sum_{n = 1}^{N} |a_n - a_{n+1}| < \infty$

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Hint. Let $S_N= \sum_{n = 1}^{N} |a_n - a_{n+1}|$. Then, since $\lim_{N\to \infty}S_N=L\in\mathbb{R}$, it follows that $(S_N)_N$ is a Cauchy sequence. Moreover $\sum_{i=m}^{n} |a_i - a_{i+1}|=S_n-S_{m-1}$.

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  • $\begingroup$ So there's two ways I'm thinking about proceeding from the hint you gave me: 1) $S_n - S_{m-1}$ is Cauchy so then $a_n$ is Cauchy or 2) $S_n - S_{m-1} < S_N$ I would have like to use the fact that $lim_{N \rightarrow \infty} S_N = L$ but it is not necessarily true that $S_n - S_{m-1} < S_N - L$ $\endgroup$ – dc3rd Feb 20 '18 at 16:13
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    $\begingroup$ I agree with the first way. $\endgroup$ – Robert Z Feb 20 '18 at 16:22
  • $\begingroup$ SO the key to this problem was recognizing that the given Series is convergent and since that is the case that implies it is Cauchy. $\endgroup$ – dc3rd Feb 20 '18 at 16:26
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    $\begingroup$ Yes. BTW, also 2) is fine $S_n - S_{m-1} \leq S_N - L$ holds because the sequence $S_N$ is increasing (the terms are non-negative). See also Martin Argerami's answer. $\endgroup$ – Robert Z Feb 20 '18 at 16:35
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Convergence of a series means precisely that the tail of the series goes to zero. That's exactly what you need.

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