Let $(a_n)$ be a sequence such that $\lim\limits_{N\to\infty} \sum_{n=1}^n |a_n-a_{n+1}|<\infty$. Show that $(a_n)$ is Cauchy.

enter image description here

So basically I am told that the sum of the difference isn't infinite. I know that to show the sequence is Cauchy, the difference between the sums must be very small ($\epsilon$). So what exactly do I have to do to answer this question? I am not having a good understanding what "new" information is giving me

  • 4
    $\begingroup$ Use the fact that the partial sums of the series form a Cauchy sequence. Then use the triangle inequality in a certain way. $\endgroup$ – wj32 Oct 7 '12 at 5:14
  • $\begingroup$ I think I know what you mean. I have my notes. But I don't understand how I can make the absolute value difference bounded to something small $\endgroup$ – Hawk Oct 7 '12 at 5:26

We will use the fact, if a series $\sum_{k=1}^{\infty} b_k$ converges, then $$ \lim_{n \to \infty}\sum_{k=n}^{\infty} b_k = 0 \,, \quad (1)\,. $$

To prove a sequence $a_n$ is a Cauchy sequence, the following has to hold $$ \lim_{n \to \infty}|a_n-a_{n+p}|=0\,, \quad \forall p\geq 1 \,.$$

Now, applying that to your problem, observe that, $$|a_n-a_{n+p}| = |(a_n-a_{n+1})+(a_{n+1}-a_{n+2})+(a_{n+2}-a_{n+3})+\dots+(a_{n+p-1}-a_{n+p})|$$ $$\implies |a_n-a_{n+p}| = \left|\sum_{k=n}^{n+p-1}(a_k-a_{k+1})\right| \leq \sum_{k=n}^{n+p-1}|a_k-a_{k+1}|\leq \sum_{k=n}^{\infty}|a_k-a_{k+1}|\,, \quad (*) $$ The last inequality follows from the fact that we are adding positive terms. Taking the limit of both sides of $(*)$ and using $(1)$, the desired result follows

$$ \lim_{n \to \infty}|a_n-a_{n+p}|=0\,, \quad \forall p\geq 1\,. $$

  • $\begingroup$ How did you jump to that last step? What happened to the limit operator? $\endgroup$ – Hawk Oct 7 '12 at 6:16
  • $\begingroup$ @jak:How do you prove a sequence is Cauchy? Review the definition. $\endgroup$ – Mhenni Benghorbal Oct 7 '12 at 6:33
  • $\begingroup$ @jak:Saying $\lim_{n \to \infty}a_n=a$ or $\forall n>N \Rightarrow |a_n-a|<\epsilon$ are equivalent. $\endgroup$ – Mhenni Benghorbal Oct 7 '12 at 6:43
  • $\begingroup$ No I am wondering how you got $a_1 - a_{n+1}$ and how you cliam that $a_{n+1} = s_n$ $\endgroup$ – Hawk Oct 7 '12 at 6:55
  • 1
    $\begingroup$ @jak: I see what you mean. Are you taking $a_n=\ln(n)$ as an example. But in this case, the condition $ \sum_{n=1}^{\infty}|\ln(n)-\ln(n+1)|<\infty$ is not satisfied. So, you can not consider this sequence. $\endgroup$ – Mhenni Benghorbal Oct 8 '12 at 3:56

Think of it this way: Let $b_n=|a_n-a_{n+1}|$. Then the statement is that $\lim_{N\rightarrow \infty}\sum_{n=1}^N b_n$ is finite, i.e. the series converges. Think about what that implies for $\sum_{n=n_1}^{n_2} b_n$ for large $n_1$ and $n_2$, and then consider how $\sum_{n=n_1}^{n_2} b_n$ compares to $|a_{n_2}-a_{n_1}|$ (which is what you're trying to get very small.)

  • $\begingroup$ Meaning one has to look at the Cauchy condition for partial sums (of the difference) is what you are saying?? $\endgroup$ – Vishesh Oct 7 '12 at 5:14

What do you know about the tails of converging series? How can you use that information to bound the distances between points in the Cauchy sequence?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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