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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 '13 at 20:25
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...such that $\lim_{N\to\infty}\sum_{n=1}^N|a_n-a_{n+1}|$...what?.. exists? –  Eckhard Jan 13 '13 at 20:25
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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 '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 '13 at 20:29
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@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 '13 at 20:38
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2 Answers

up vote 5 down vote accepted

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. –  B. S. Jan 16 '13 at 14:25
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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 '13 at 20:36
    
@MattN. For the record, if you made $|a_{n+1}-a_n|<b_n$ for some summable $b_n$, then you could argue $a_n$ is Cauchy. –  Pedro Tamaroff Apr 1 '13 at 21:55
    
@PeterTamaroff Why summable? Letting $b_n = 1/n$ one could argue that $a_n$ is Cauchy. –  Matt N. Apr 2 '13 at 6:41
    
@MattN. No. Say $m<n$, you'd make $$|a_m-a_n|\leq |a_m-a_{m+1}|+|a_{m+1}-a_{m+2}|+\cdots+|a_{n-1}-a_n|=\sum_{k=m}^{n-1} |a_{k}-a_{k+1}|<\sum_{k=m}^{n-1}b_k$$ Since $b_k$ is summable, you get $<\epsilon$ at the end for $n,m>N$. –  Pedro Tamaroff Apr 2 '13 at 6:47
    
@PeterTamaroff Sorry, I read $|a_m - a_n|<b_n$ and assumed $m \geq n$. –  Matt N. Apr 2 '13 at 7:29
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