Let’s Get Real… Analysis (Part 10): Cool Stuff with Cauchy Sequences

What does it mean for a sequence to be a Cauchy sequence? Why are Cauchy sequences important? Great questions! Hopefully by the end of this article you can answer these for yourself!

The Need for More

Recall back in Let’s Get Real… Analysis (Part 2): Sequence Limits we learned that a sequence converges to some limit , if for all there exists an such that for all which is a mouthful! Or, for those who love symbols:

$a_n\rightarrow \alpha \;\;\;\mathrm{if}\;\;\;\forall\varepsilon>0,\,\exists N\in\mathbb{N}\,:\,|a_n - \alpha| < \varepsilon,\; \forall n>N.$

On the off chance that you haven’t learned this definition you can continue reading this article, however, you will most likely get more out of this article by going back to part 2 to get more comfortable with this messy scramble of symbols. But, for those who remember and are comfortable with this definition let me ask you a question that we have asked before.

What would you do if you wanted to know that $(a_n)$ converges, but you had no idea what it’s limit might be?

Trying to give a proof by definition isn’t very helpful in this situation, since we need to show that $|a_n - \alpha| < \varepsilon$ we need to know $\alpha.$ We’ve answered this question when $a_n$ is a monotone sequence, and the answer came in the form of the monotone convergence theorem. However, we don’t have a way to answer this question generally. This is our motivation for finding some other criteria that we can use to check to see if $(a_n)$ converges without relying on what $\alpha$ in order to use it. Enter the genius of Cauchy.

What are Cauchy Sequences?

We’re given some sequence, $(a_n),$ and our task is to determine whether or not $(a_n)$ converges without needing to try to determine its limit beforehand.

What can we do?

Well, suppose $(a_n)$ did converge to some limit $\alpha.$ Then we would know that the elements $a_n$ get closer and closer to $\alpha$ (as you increase $n$ ). It follows that the elements in $(a_n)$ are also getting closer to one another! For example, $a_n$ and $a_{n+1}$ must get closer and closer together as they (both) approach $\alpha$ . This seems promising, we are no longer comparing $a_n$ to $\alpha,$ we are comparing $a_n$ and $a_{n+1}$ . However, let’s not be fooled, the logic we went through is the converse of what we want and just because a statement is true, the converse of that statement need not be true.¹

What we said: When $(a_n)$ converges, then $(a_{n+1}-a_n) \rightarrow 0$ . (I leave it as a challenge to prove this!)

What we want to be true: If $a_{n+1}-a_n \rightarrow 0$ , then $(a_n)$ converges.

The question is now, is what we want to be true indeed true?

Take a moment to ponder this query…

If you think it’s true, can you prove it? If you think it’s false, can you find a counter example where $a_{n+1}-a_n \rightarrow 0$ and $(a_n)$ still diverges?

If you think that what we want to be true is true, you’re very close, but we need to modify what we want to be true a little. If you thought it was false, could you find a counter example? I’ll put a counter example in the pull down menu below, so that way I don’t spoil the answer for you!

A counter example to what we want to be true

Let $(a_n) = \sqrt{n}$ . Then it can be shown that $\sqrt{n+1} - \sqrt{n} \rightarrow 0$ and $a_n \rightarrow \infty$ . I leave it as a challenge to prove these statements (you’re welcome!) But let me give you a hint: use the formula $(a^2 - b^2) = (a+b)(a-b)$ .

$\square$

Ok, back to the drawing board. It was a good idea to start comparing different elements $a_n$ and $a_{n+1}$ from our sequence. The problem is that there are sequences that diverge who have consecutive elements that get closer and closer to one another. How might we fix this issue?

We shouldn’t be comparing only consecutive elements from $(a_n),$ we should be comparing all elements from $(a_n)$ to each other! Since $a_n\rightarrow \alpha$ we know that, for any $\varepsilon>0,$ there is some $N$ such that for all In other words, for $n>N$ and $m>N$ we have,

$|a_n - a_m| = |a_n - \alpha - (a_m - \alpha)| \\ \textit{ }\qquad \;\;\;\;\;\;\leq |a_n - \alpha| + |a_m - \alpha| \\ \textit{ }\qquad \;\;\;\;\;\; < \frac{\varepsilon}{2}+ \frac{\varepsilon}{2} \\ \textit{ }\qquad \;\;\;\;\;\; =\varepsilon.$

where we used the triangle inequality and that $n,m>N$ .

This new guess works just as well as our last, in that, if $a_n\rightarrow \alpha$ then $|a_n - a_m| < \varepsilon$ for all $n,m>N$ . But again, we still haven’t shown that the converse is true: If $|a_n - a_m| < \varepsilon$ for all $n>N$ and $m>N,$ then $(a_n)$ converges. As we will soon see, the converse in the previous sentence is indeed true and is known as Cauchy’s criteria! But, before that a quick definition.

Definition (Cauchy Sequences): Let $(a_n)$ be a sequence of real numbers. If $(a_n)$ has the property that, for all there exists an such that for all then we call $(a_n)$ a Cauchy sequence. In symbols,

$\text{ }\;\;\;\;\;(a_n)\;\mathrm{is}\;\mathrm{called}\;\mathrm{a}\;\mathrm{Cauchy}\;\mathrm{sequence}\;\mathrm{if}\;\\\forall\varepsilon>0,\,\exists N\in\mathbb{N}\,:\,|a_n - a_m| < \varepsilon,\; \forall n,m>N.$

Notice that a sequence being Cauchy makes no reference to any limit values. Actually, being Cauchy makes no reference to convergence at all! This is its utility.

The Cauchy – Convergence Connection

Let me ask you. Which do you think is true:

If a sequence is Cauchy, then it converges. Or,
If a sequence converges, then it’s Cauchy.

If you guessed i. was correct good job! It is! However, if you guessed ii. was correct, then also good job! Because i. and ii. are correct Now, if you said both i. and ii., then you are right on the money! Good job!

Combining both i. and ii. into one statement is known as Cauchy’s Convergence Criteria and is the last thing we will

Theorem (Cauchy’s Convergence Criteria): Let $a_n$ be a sequence of real numbers. Then $(a_n)$ converges to some limit if and only if $(a_n)$ is Cauchy.

Our proof for the forward direction will mimic what we discussed in the motivation section. Proving that: Cauchy implies convergence is a little trickier!

Proof: (Click in the Discovery)

Let $(a_n)$ be a sequence of real numbers. We will break up both directions.

Forward: If $(a_n)$ converges to some limit, then $(a_n)$ is Cauchy.

Let $\varepsilon>0$ and assume that $(a_n)$ converges to $\alpha$ and we will show that $(a_n)$ is Cauchy. Using the definition of convergence we know that there exists some $N$ such that $|a_n - \alpha| < \varepsilon/2$ for all $n>N$ . Thus, for all $n>N$ and $m>N$ we have

$|a_n - a_m| = |a_n - \alpha - (a_m - \alpha)| \\ \textit{ }\qquad \;\;\;\;\;\;\leq |a_n - \alpha| + |a_m - \alpha| \qquad\mathrm{by}\;\mathrm{the}\; \mathrm{triangle}\;\mathrm{inequality}\\ \textit{ }\qquad \;\;\;\;\;\; < \frac{\varepsilon}{2}+ \frac{\varepsilon}{2} \\ \textit{ }\qquad \;\;\;\;\;\; =\varepsilon.$

This shows that $a_n$ is Cauchy and concludes the forward direction.

$\square_{\mathrm{Forward}}$

Backward: If $(a_n)$ is Cauchy, then $(a_n)$ converges to some limit.

Let $\varepsilon>0$ and assume that $(a_n)$ is Cauchy. I would argue that this is the more challenging direction to prove. How might we show that $(a_n)$ converges? The problem is that we need to know what $(a_{n})$ ‘s limit is prior to being able to prove it converges, and $(a_{n})$ being Cauchy doesn’t give us this limit value. Our first task, then, is to determine a possible value that $(a_{n})$ might converge to.

Since we know that $(a_n)$ is Cauchy, there exists an such that for all . This means that,

$-1 + a_{N_1 +1}< a_n < 1 + a_{N_1 +1},$

for all $n>N_1.$ It follows that $a_n$ is bounded for But, since there is only a finite number of elements from $(a_n)$ with $n\leq N_1,$ (there are $N_1$ of them) we know that those values are bounded by some maximum $a_M$ . Taken together we have,

$|a_n| \leq \max{\Big\{ |a_1|\,,\,\cdots\,,\,|a_2|\,,\, |1 + a_{N_1 +1}|\,,\,|1 - a_{N_1 +1}| \Big\} }$

and hence $(a_n)$ is bounded. Which puts us in the position to use the Bolzano-Weierstrass Theorem from last time in order to conclude that $(a_n)$ has a convergent subsequence $(a_{n_k})$ . Let’s denote the value that $(a_{n_k})$ converges to by $\alpha,$ hence $a_{n_k} \rightarrow \alpha$ as $k\rightarrow \infty.$

Recall from, Let’s Get Real… Analysis (Part 8): Superb Subsequence, that a sequence converges to $\alpha$ if and only if every subsequence converges to $\alpha.$ So, if $(a_n)$ is going to converge to anything, it’s going to be $\alpha.$ Therefore, we want to prove that $a_n \rightarrow \alpha,$ and so we want to find some such that,

$\;\;\;|a_n - \alpha| < \varepsilon,$

for all First, we know that there is some $K\in\mathbb{N}$ such that,

$|a_{n_k} - \alpha| < \frac{\varepsilon}{2}.$

for all $k>K.$ Since $(a_n)$ is Cauchy we also know there is some $M\in\mathbb{N}$ so that way we have

$|a_n - a_{n_k}| < \frac{\varepsilon}{2}.$

for all $n>M$ and $k>M.$ This motivates us to let $N = \max{ \{K+1,M+1 \}}.$ (The “+ 1” are there to ensure that $N>M$ and $N>K$ without needing to put a “+ 1” in a subscript in the following equations.) Thus,

$|a_n - \alpha| = |a_n - a_{n_N} + a_{n_N} - \alpha| \\ \textit{ }\qquad \;\;\;\; \leq |a_n - a_{n_N}| + |a_{n_N} - \alpha| \\ \textit{ }\qquad \;\;\;\; =\frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ \textit{ }\qquad \;\;\;\; = \varepsilon$

for all $n>N.$ We have shown that $(a_n)$ converges and conclude the proof.

$\square$

Stay frosty!

When I first learned Cauchy’s criterion, I found it incredibly ingenious! Having the ability to show that a sequence converges without having to put in the effort of finding it’s limit is so useful! Especially when you don’t have a calculator to help you determine the value of the limit. There are countless problems out there that you can now tackle, go forth and prove that sequences converge!!!!

Oh, one last thing. Cauchy sequences play a crucial role in more advanced analysis; they are used when we have more abstract metric spaces than the real numbers. Pay no attention to that statement if you are unfamiliar with metric spaces! I just thought we should plant the seed for your next steps!

Be Kind. Be Curious. Be Compassionate. Be Creative.

And Have Fun!

Footnotes:

For example, take the statement: If it rains, I bring an umbrella to work. The converse is, if I bring an umbrella to work, then it’s raining. Which doesn’t have to be true! Maybe I bring an umbrella every day, even sunny, cloudless days! ↩︎

A Kick in the Discovery