# Show that a Sequence is Cauchy.

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

My work: $$\lim_{N\to \infty} \sum_1^N |a_n - a_{n+1}| = |a_1 - a_2| + \cdots+|a_{N-1} - a_N| \ge |a_1| + |a_3|+\cdots+ |a_{2n+1}|$$ for $n \ge 1$ Therefore since $\infty \gt \sum_1^N |a_n - a_{n+1}| \ge \sum_1^{2n+1} |a_n|$ and |$a_n$| is positive, it is finite. Therefore the subsequence of $|a_n|$ for odd positive integers converges, so by Bolzano-Weierstraß theorem, the sequence is bounded, therefore the sequence is cauchy.

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I think you're missing something...$a_n$ is a sequence such that the stated limit is finite, or...? – icurays1 Jan 13 at 20:25
...such that $\lim_{N\to\infty}\sum_{n=1}^N|a_n-a_{n+1}|$...what?.. exists? – Eckhard Jan 13 at 20:25
Your inequality isn't true if we let $a_n=1$ for all $n$. That is, where you say $\sum |a_n-a_{n+1}|\geq\sum |a_{2n-1}|$. – Clayton Jan 13 at 20:26
It is good that you show your own work, but it is full of mistakes. Not to mention that those guys were not called Balzano and Wierstrauss. – TMM Jan 13 at 20:29
@ReubenPereira: What you have in your comment is correct, but you lose the inequality if you drop $|a_{n+1}|$ as my example shows. – Clayton Jan 13 at 20:38

Since $\sum_{n=1}^\infty|a_n-a_{n+1}|<\infty$, we can say that for any $\varepsilon>0$, there exists an $N\in\Bbb N$ such that $$\sum_{n=N+1}^\infty|a_n-a_{n+1}|<\varepsilon.$$ Now, by definition, a sequence is Cauchy if given $\varepsilon>0$, there exists $N>0$ such that $|a_k-a_m|<\varepsilon$ for $k,m>N$. But note that $$|a_k-a_m|\leq\sum_{n=k}^m|a_n-a_{n+1}|\leq\sum_{n=k}^\infty|a_n-a_{n+1}|<\varepsilon.$$ Thus, you have a Cauchy sequence.

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 +1 Nice approach, Clayton. – Babak S. Jan 16 at 14:25

Hint: For $m <n$ we have

$$\left|a_n-a_m \right| \leq \left|a_n-a_{n-1} \right|+\left|a_{n-1}-a_{n-2} \right|+..+\left|a_{m+1}-a_m \right|=s_n-s_{m-1}$$

where $s_n$ is the partial sum of your series.

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 Thanks a lot for your comment! – Matt N. Jan 13 at 20:36 @MattN. For the record, if you made $|a_{n+1}-a_n|N$. – Peter Tamaroff Apr 2 at 6:47 @PeterTamaroff Sorry, I read \$|a_m - a_n|
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