Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
...such that $\lim_{N\to\infty}\sum_{n=1}^N|a_n-a_{n+1}|$...what?.. exists? – Eckhard Jan 13 '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 '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
@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
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.

share|cite|improve this answer
+1 Nice approach, Clayton. – Babak S. Jan 16 '13 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.

share|cite|improve this answer
Thanks a lot for your comment! – Rudy the Reindeer 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. – Rudy the Reindeer 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 $m \geq n$. – Rudy the Reindeer Apr 2 '13 at 7:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.