Does $|a_n-a_{n+1}|\to 0$ imply $(a_n)$ is Cauchy?

Question

My textbook has this problem as a kind of "concept check", where one is supposed to find a counterexample to the following statement:

A sequence of real numbers is cauchy iff.

$$ \forall \epsilon>0, \, \exists N \in \mathbb{N}, \, \forall n \geq N: |a_n-a_{n+1}|< \epsilon $$

I know that a sequence of real numbers is cauchy if

$$ \forall \epsilon>0, \, \exists N \in \mathbb{N}, \, \forall n,m \geq N: |a_n-a_m|< \epsilon $$

Finding such a counterexample is probably trivial, but I haven't been able to think of one.

Answer 1

Take $a_n=\sum_{i=1}^n\frac{1}{i}$. Then $a_{n+1}-a_n=\frac{1}{n+1}<\epsilon$ for large enough $n$. This drives home the point that you need bounds on $n,m$ simultaneously. Since the harmonic sum diverges, it follows that $|a_n-a_m|$ will grow unbounded if you fix either $n$ or $m$ and vary the other.

You can also find examples of bouned sequences $a_n$. For example define $a_n:=\exp(i\left\{\sum_{i=1}^n\frac{1}{j}\right\})$, where the brackets denote the fractional part, so that $a_n$ loops around the unit-circle perpetually and hence has no limit.

Answer 2

$a_n = \ln n$ will do the job.

Answer 3

My go-to counterexample here is to take partial sums of the harmonic series. That is, $$a_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$$ Obviously the difference between $a_{n-1}$ and $a_n$ is just $\frac{1}{n}$, which goes to $0$, but this sequence does not converge.

Answer 4

The conceptual "gothcha" is that you should realize that the terms $a_i$ can get very close to each other but cumulatively on the large scale remain unbounded. Conceptually that should be easy to see that it is possible. But coming up with an example may be hard. There are a few famous ones but what to do if you just don't think of them.

(Truth is, of all the experienced mathematicians, say "Come on, don't you know one? What this obvious sequence every one has seen", probably only one in five figure this out on their own.)

Can we visualize this. Imagine a sequence where the terms get closer together sequentually but never converge. What would they look like if you graphed them? The terms will get so close together that the graph would barely be increasing ... and you it would. At one point it might take 10 terms to get an increase of 1 unit. And latter it might to a hundred, a thousand, a million terms to get an increase of 1 unit.

Does this remind you of any graph or function? (You can have a sequence where $a_n = f(n)$ for some real function, but only evaluated at the integers.) There is one very basic and fundamental function that behaves exactly like these and for very good reason.

So hint: Let $a_0= 0$ and $a_1= 1$ and then it takes more to get to $2$. $a_{10} = 2$ and then it takes even more to get to $3$... $a_{100} = 3$ and it then starts getting real slow. $a_{1000} =4$ and $a_{10,000} = 5$ and $a_{100,000} = 6$. Do you see how I'm forcing the issue? Do you think you can come up with some more formally defined sequence that acts the same way?

Answer 5

$$0,1/2,1,1/2,0,1/3,2/3,1,2/3,1/3,0,1/4,2/4,3/4,1,3/4,2/4,1/4,0,\dots $$