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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.

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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

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Use the fact that the partial sums of the series form a Cauchy sequence. Then use the triangle inequality in a certain way. – wj32 Oct 7 '12 at 5:14
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 – Hawk Oct 7 '12 at 5:26
up vote 3 down vote accepted

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

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

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How did you jump to that last step? What happened to the limit operator? – Hawk Oct 7 '12 at 6:16
@jak:How do you prove a sequence is Cauchy? Review the definition. – Mhenni Benghorbal Oct 7 '12 at 6:33
@jak:Saying $\lim_{n \to \infty}a_n=a$ or $\forall n>N \Rightarrow |a_n-a|<\epsilon$ are equivalent. – Mhenni Benghorbal Oct 7 '12 at 6:43
No I am wondering how you got $a_1 - a_{n+1}$ and how you cliam that $a_{n+1} = s_n$ – Hawk Oct 7 '12 at 6:55
@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. – 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.)

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Meaning one has to look at the Cauchy condition for partial sums (of the difference) is what you are saying?? – 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?

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