# Cauchy Sequences: the Importance of Getting Close

I am an analyst at heart, despite my recent set of algebra posts. Augustin Louis Cauchy can be argued as one of the most influential mathematicians in history, pioneering rigor in the study of calculus, almost singlehandedly inventing complex analysis and real analysis, though he also made contributions to number theory, algebra, and physics.

One of the fundamental areas he studied was sequences and their notion of convergence. Suppose I give you a sequence of numbers, and ask you what happens to this sequence if I kept appending terms forever? Would the path created by the sequence lead somewhere?

Let’s start with a nice, basic sequence, and assume we are living on the space of real numbers:

s = (1, 1/2, 1/3, 1/4, 1/5, 1/6,\ldots)That tuple notation denotes an ordered *sequence* of stuff^{1}. The individual elements are *terms* of the sequence. The first term in s above is 1, called s_{1}. So s_{6} = 1/6.^{2}. The ellipses tell you that there is no end to the sequence; that it goes on forever.

## First question: what’s the pattern?

If we’re going to study sequences, we need to know how to refer to one in general. What’s the 100th term of the sequence I showed you? The millionth? The general nth term of the sequence is a function of its index.^{3}.

In our sequence above, s_{1} = 1, s_{2} = 1/2, s_{3} = 1/3…notice a pattern? Can you see how to use the index to generate terms of the sequence?

s_{n} = 1/n; that is, the nth term of the sequence is obtained by dividing 1 by the index n. So we can actually represent the sequence a bit shorter: s_{n} = \left(\frac{1}{n}\right)This tells us how to generate the entire sequence, term by term. Now we can speak about a generic sequence a with terms a_{n}, where n is the index.

## Studying sequences – the Cauchy property

One of the biggest questions we can ask of a sequence is regarding *convergence. *Where do the terms lead, if anywhere? Convergence theory of both sequences and series (the sum of a sequence) is quite a rabbit hole to dive down, but in short, we want to know if there is some “destination” of the sequence. Proving convergence is required; it’s not enough to just say where they go and call it good. For now, though we’re going to play with a couple sequences to see if we can get an intuition of what happens.

### (1, 1/2, 1/3, 1/4, 1/5,\ldots)

This one is a sequence of fractions, given by the general term s = \left(\frac{1}{n}\right). Where do you think these terms lead? If you guessed 0, then you are right.^{4}.

Let’s try another one.

### (1,0,1,0,1,0,\ldots)

This sequence alternates between 0 and 1 forever. So where does it lead? In this case, nowhere. It’s just an eternal pong match between 0 and 1.

One last example:

### (1, 1, 2, 3, 5, 8, 13, \ldots)

This one is the famous Fibonnaci sequence, as I’m sure most of you recognized. The next term is generated by adding the two previous terms. So if we kept generating terms of the sequence forever, where would we go? The terms just keep getting bigger and bigger, so we would never actually converge to anything finite. We’d be walking in this Fibonnaci tunnel forever, never actually reaching some number we would converge to.

All three of these are sequences, but only one of them has a very useful and famous property- the **Cauchy Property**.

In a nutshell, a sequence that has the Cauchy property (or is *Cauchy*), has terms that get arbitrarily close after a certain point. Put formally, for every tiny \epsilon > 0 there exists some number N such that for any two indices m and n that are greater than this N,

Let’s take this apart to understand what this definition means.

### For every \epsilon > 0

This means that I should be able to pick any number, no matter how tiny, and there is a corresponding natural number N that depends on this epsilon. I could also write N_{\epsilon} to explicitly show this dependence. Let’s work with our sequence s = \left(\frac{1}{n}\right). If you picked \epsilon = \frac{1}{4}, then after which index are any two terms less than 1/4 apart? Is it N = 2?

Well, the difference between any two subsequent terms are obviously going to get smaller, so the largest difference between two subsequent terms would be between 1/2 and 1/3^{5}.

But wait. Let’s go back to the definition. It says that we have to find an N such that * for any two* terms afterwards. That means we can’t just check the difference between terms next to each other. We have to know that it’s true for any two, such as the second and fifth, or fourth and twentieth. All the terms have to be bunching together after a certain point.

So in our case, for example, 1/2 – 1/5 = 3/10 > 1/4, so N = 2 isn’t our N.

### How do we actually show a sequence is Cauchy?

Testing every possible \epsilon and finding its corresponding N is going to be impossible. It’s a pain just to do it for \epsilon = \frac{1}{4} above. This means we do need to attack this problem abstractly. Showing a sequence is Cauchy is actually pretty formulaic, because we have to satisfy a definition. We’ll walk through it on our sequence s = \left(\frac{1}{n}\right), because this exercise forces us to really understand the definition.

First, the definition says “for every \epsilon > 0“. That means we need to grab one – a generic one. So let’s do that. We say: **Let \epsilon > 0** to start out. Now we have one. The goal is to find the N in terms of our generic \epsilon such that any two terms of s = \left(\frac{1}{n}\right) with indices greater than this N have a difference no larger than our chosen (but unknown) \epsilon.

So let’s set up what we need: \left|\frac{1}{n}-\frac{1}{m}\right| < \epsilon. We’ll need to just play with it algebraically to get somewhere.

\begin{aligned}\left|\frac{1}{n}-\frac{1}{m}\right| &= \left|\frac{m-n}{mn}\right|\end{aligned}All I did in the line above was combine the fraction. Let’s see here, if I subtract n from m in the numerator, that’s less than not subtracting anything at all. That means that

\begin{aligned}\left|\frac{1}{n}-\frac{1}{m}\right| &= \left|\frac{m-n}{mn}\right|\\&<\left|\frac{m}{mn}\right|\\&= \frac{1}{n}\end{aligned}Why did I do this? I want to bound what looked icky by something that looks nicer to work with.^{6}. Now, how do we ensure that \frac{1}{n} is less than an \epsilon we don’t know? This is where our N comes in. What should N be in terms of our \epsilon to guarantee this?

If N = \frac{1}{\epsilon}, and n > N, then \frac{1}{n} < \frac{1}{N} = \frac{1}{1/\epsilon} = \epsilon.

That means that for N = \frac{1}{\epsilon}, and for any two m,n that are greater than N \begin{aligned}\left|\frac{1}{n}-\frac{1}{m}\right| &= \left|\frac{m-n}{mn}\right|\\&<\left|\frac{m}{mn}\right|\\&= \frac{1}{n}\\&<\epsilon\end{aligned}

We’re done! We took a generic \epsilon and found the corresponding N that ensured any two terms with an index greater than N are no further than \epsilon apart. Because we didn’t specify anything, we showed that this is true no matter what \epsilon you picked. Therefore, we conclude that the sequence s = \left(\frac{1}{n}\right) is a Cauchy sequence.

[*Some notes: I want to point out here that a Cauchy sequence doesn’t have to be super neat. As long as it “calms down” after a certain finite point, it can be wild and crazy. For example, the first 10 terms of a sequence can be 1,000,000, and then from the 11th term onward be something like \left(\frac{1}{n^{2}}\right). When we walk about sequences, we actually don’t care at all about the first bits of a sequence. It’s the tail we care about almost always.]*

## So what does this give us?

The Cauchy property actually yields quite a few things that can help us when we study convergence of both sequences and series. Here are a few things we can prove if we know a sequence is Cauchy:

**(1) Every Cauchy sequence of real or complex numbers is bounded. **

**(2) A Cauchy sequence that has a convergent subsequence is itself convergent. **

A subsequence is a sequence made from selecting certain terms of the sequence to make a new one, like all the odd terms, or all the evens, or every third one, etc. This can reduce the problem of showing convergence of a complicated sequence if we can find a subsequence that leads somewhere.

**(3) Every Cauchy sequence of real numbers converges. **

This is actually a very specific case of a more general statement, that every Cauchy sequence in n-dimensional real space is convergent. This is extremely helpful when we want to show a sequence converges, but we can’t really figure out what the limit might be by intuition.

## Conclusion

This post was meant to give a slight taste of mathematical analysis in the form of studying sequences of numbers with the Cauchy property. Of course, we can get more general than this, and study Cauchy sequences of anything we like: functions, groups, vectors, and more abstract objects than that. We’ll tackle more concepts in analysis in future posts.

This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.

#### Footnotes

- We can make sequences of anything we want to: matrices, numbers, functions, or even more general objects. We’ll stick with numbers for this discussion
- Some people start indexing terms at 0. Either one is fine, just be aware of which one you have or choose.
- Actually, a sequence is a function that maps an index, typically the natural (counting) numbers onto some other space
- Again, we can prove this formally, but I want to develop a sense of intuition here without getting bogged down in epsilons and scaring people off
- We left out the first term, because we want to see if after the second term, all terms are less than 1/4 apart.
- In my experience, analysis is a game of bounding arguments. It’s a neat way to sidestep some really nasty arithmetic or calculus.