2.9 The Cauchy criterion

As we have seen, the monotone convergence theorem is useful because it allows us to show a limit exists without knowing the precise value of the limit (as opposed to, say, working directly with Definition 2.23). However, it has one major drawback: it only works for monotone sequences! There are plenty of sequences which are not monotone, but nevertheless are convergent: for a good example, consider $(\sin(n)/n)_{n\in\mathbb{N}}$ . However, there is a very powerful tool for deducing the existence of limits which does not rely on monotonicity: the Cauchy criterion.

Definition 2.77.

A sequence $(a_{n})_{n\in\mathbb{N}}$ of real numbers is Cauchy if for every $\varepsilon>0$ there exists some $N\in\mathbb{N}$ such that

|a_{n}-a_{m}|<\varepsilon\qquad\text{for all $n$, $m>N$.}

Example 2.78.

Consider the sequence $(a_{n})_{n\in\mathbb{N}}$ given by $a_{n}:=1/n$ for all $n\in\mathbb{N}$ . We claim $(a_{n})_{n\in\mathbb{N}}$ is Cauchy. To see this, let $\varepsilon>0$ be given and choose $N:=\lceil 1/\varepsilon\rceil$ . If $n$ , $m>N$ with $n>m$ , say, then

|a_{n}-a_{m}|=\frac{1}{m}-\frac{1}{n}=\frac{n-m}{nm}\leq\frac{n}{nm}=\frac{1}{% m}<\frac{1}{N}<\varepsilon.

Hence, by definition, $(a_{n})_{n\in\mathbb{N}}$ is Cauchy.

Intuitively, a sequence is Cauchy if for large values of $n\in\mathbb{N}$ the terms $a_{n}$ all cluster together. This is certainly true of a convergent sequence, where the terms eventually cluster around the limit value. The following theorem makes this observation precise.

Theorem 2.79.

If $(a_{n})_{n\in\mathbb{N}}$ is a convergent sequence of real numbers, then $(a_{n})_{n\in\mathbb{N}}$ is Cauchy.

Proof.

If $(a_{n})_{n\in\mathbb{N}}$ is convergent, then there exists some $a\in\mathbb{R}$ such that $\lim_{n\rightarrow\infty}a_{n}=a$ . From the definition of convergence, for any given $\varepsilon>0$ , there exists $N\in\mathbb{N}$ such that $|a_{n}-a|<\varepsilon/2$ for all $n>N$ . Using the triangle inequality,

|a_{n}-a_{m}|=|a_{n}-a+a-a_{m}|\leq|a_{n}-a|+|a-a_{m}|<\varepsilon/2+% \varepsilon/2=\varepsilon

whenever $n$ , $m>N$ , hence the sequence is Cauchy. ∎

A very important result is that the converse of Theorem 2.79 is also true. This observation relies heavily on the completeness axiom.

Theorem 2.80 (Cauchy criterion).

Let $(a_{n})_{n\in\mathbb{N}}$ be a sequence of real numbers. Then $(a_{n})_{n\in\mathbb{N}}$ is Cauchy if and only if $(a_{n})_{n\in\mathbb{N}}$ is convergent.

The Cauchy criterion is a powerful result with many important consequences. In particular, like the monotone convergence theorem, the Cauchy criterion allows us to deduce the existence of a limit of a sequence without having to know the precise value of the limit beforehand. Consequently, it is very useful in existence proofs. However, unlike the monotone convergence theorem, the Cauchy criterion does not involve any additional hypotheses such as monotonicity and therefore can be applied more widely. We shall see many applications of this theorem in the next section when we investigate series.

Proof (of Theorem 2.80).

We have already shown that any convergent sequence is Cauchy; it remains to prove the converse. Supposing that $(a_{n})_{n\in\mathbb{N}}$ is Cauchy, we break the proof into steps.

Step 1. First we show that a Cauchy sequence is bounded. Indeed, let $\varepsilon:=1$ (or any fixed, positive value). Then there exists some $N\in\mathbb{N}$ such that $|a_{n}-a_{m}|<1$ for all $n$ , $m>N$ . In particular, by the triangle inequality, we have $|a_{m}|<1+|a_{N+1}|$ for all $m>N$ and so

|a_{n}|\leq\max\{|a_{1}|,|a_{2}|,\dots,|a_{N}|,|a_{N+1}|+1\}\qquad\text{for % all $n\in\mathbb{N}$.}

It therefore follows that the sequence $(a_{n})_{n\in\mathbb{N}}$ is bounded.

Step 2. We now find a convergent subsequence. Indeed, by the Bolzano–Weierstrass Theorem, since $(a_{n})_{n\in\mathbb{N}}$ is bounded by Step 1, there must exist a convergent subsequence, say $(a_{n_{k}})_{k\in\mathbb{N}}$ with $a_{n_{k}}\to a$ as $k\to\infty$ for some $a\in\mathbb{R}$ .

Step 3. Finally, we show that $a_{n}\to a$ . Indeed, let $\varepsilon>0$ be given. Since $(a_{n})_{n\in\mathbb{N}}$ is Cauchy, we can choose some $N\in\mathbb{N}$ such that

|a_{n}-a_{m}|<\frac{\varepsilon}{2}\quad\text{for all $n$, $m>N$.}

Also, since $a_{n_{k}}\to a$ as $k\to\infty$ , we can find some $K\in\mathbb{N}$ such that

|a_{n_{k}}-a|<\frac{\varepsilon}{2}\quad\text{for all $k>K$.}

Fix an index $k\in\mathbb{N}$ such that $k>K$ and also $n_{k}>N$ . Then

|a_{n}-a|\leq|a_{n}-a_{n_{k}}|+|a_{n_{k}}-a|<\frac{\varepsilon}{2}+\frac{% \varepsilon}{2}=\varepsilon\qquad\text{for all $n>N$.}

Thus, by the $\varepsilon$ - $N$ definition of a limit, $a_{n}\to a$ as $n\to\infty$ . ∎