4.5 Cauchy Sequences

Our goal was to develop tools that allow us to determine if a sequence is convergent without knowing what the limit of sequence is. We know that if a sequence is monotone and bounded then it is convergent, and that if a sequence is a subsequence of a bounded sequence, then it is convergent. But what if a sequence does not fulfill these properties? Our current definition of convergence requires knowledge of the limit in order to prove that a sequence is convergent, so how about we create a new definition of a convergent sequence where we don’t need to know what the limit is. The intuition for this defintion is that we want to express mathematically that for a sequence to converge the terms have to get closer and closer together as $n\to\infty$ . We will name this new type of convergent sequence Cauchy after the 19th-century French mathematician Augustin-Louis Cauchy.

Definition 4.5.

Let $(x_{n})_{n}$ be a sequence of real numbers. Then $(x_{n})_{n}$ is a Cauchy sequence if

\forall\varepsilon>0,\ \exists N\in\mathbb{N}\ :\ n,m\geq N\implies\lvert x_{m% }-x_{n}\rvert\leq\varepsilon.

(4.42)

What this defintion is encoding for us is that for any small $\varepsilon$ , there exists a step in the sequence such that from that step onwards, all terms of the sequence are $\varepsilon$ -close to each other. It is actually quite important that we specify ‘all terms’ as we want this definition to be as close as possible to our original definition of convergence and if we say that, for example, only consecutive terms must get $\varepsilon$ -close to each other then we admit some sequences which we would not consider convergent under the original definition. For example, the sequence $(x_{n})_{n}=(\sqrt{n})_{n}$ . Successive terms get arbitrarily close to each other ( $x_{n+1}-x_{n}=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2% \sqrt{n}}$ ) but not all others (the terms still get arbitrarily large as $n\to\infty$ so for any index $n$ and distance $M$ we can always find a large enough index $m$ such that $x_{m}-x_{n}>M$ ), and thus the sequence is not Cauchy.

Example 4.10.

Show that $\left(\frac{1}{n}\right)_{n}$ is a Cauchy sequence.
Let $\varepsilon>0$ and choose $N>\frac{2}{\varepsilon}$ . Then for $m,n\geq N$ we have

\lvert x_{m}-x_{n}\rvert=\left\lvert\frac{1}{m}-\frac{1}{n}\right\rvert\leq% \frac{1}{m}+\frac{1}{n}\leq\frac{1}{N}+\frac{1}{N}=\frac{2}{N}<\varepsilon.

(4.43)

Example 4.11.

Show that $\left(\frac{n}{1+n}\right)_{n}$ is a Cauchy sequence.
Let $\varepsilon>0$ and choose $N>\frac{1}{\varepsilon}$ . Then for $m\geq n\geq N$ we have

	$\displaystyle\lvert x_{m}-x_{n}\rvert$	$\displaystyle=\left\lvert\frac{m}{1+m}-\frac{n}{1+n}\right\rvert=\left\lvert% \frac{m-n}{(1+m)(1+n)}\right\rvert$		(4.44)
		$\displaystyle\leq\frac{m-n}{mn}=\frac{1}{n}-\frac{1}{m}\leq\frac{1}{n}\leq% \frac{1}{N}<\varepsilon.$		(4.45)

Example 4.12.

Show that the sequence given by $x_{n}=1+\frac{1}{2}+\dots+\frac{1}{n}$ is not Cauchy.
To prove this, we will show that the sequence fulfills the negation of the Cauchy sequence definition ( $\exists\varepsilon>0$ such that $\forall N\in\mathbb{N}\ \exists m,n\geq N$ such that $\lvert x_{m}-x_{n}\rvert>\varepsilon$ ).
Choose $\varepsilon<\frac{1}{2}$ , let $N\in\mathbb{N}$ , and choose $n=N$ and $m=2N$ . Then

	$\displaystyle\lvert x_{m}-x_{n}\rvert=\lvert x_{2N}-x_{N}\rvert$		(4.46)
	$\displaystyle=\left\lvert\left(1+\frac{1}{2}+\dots+\frac{1}{N}+\frac{1}{N+1}+% \dots+\frac{1}{2N}\right)-\left(1+\frac{1}{2}+\dots+\frac{1}{N}\right)\right\rvert$		(4.47)
	$\displaystyle=\frac{1}{N+1}+\dots+\frac{1}{2N}\geq\frac{1}{2N}+\dots+\frac{1}{% 2N}=N\frac{1}{2N}=\frac{1}{2}>\varepsilon.$		(4.48)

4.5.1 Cauchy and Convergent Sequences are Equivalent

Notice that in these examples, the sequences that are Cauchy are also convergent, and the sequence that is not Cauchy is not convergent. This correspondence turns out to be general and in fact it is an important result that the criteria for being convergent and being Cauchy are equivalent.

Theorem 4.9

Let $(x_{n})_{n}$ be a sequence of real numbers. Then $(x_{n})_{n}$ is a convergent sequence if and only if it is also a Cauchy sequence.

Proof.

(Forward direction $\implies$ ): We will make use of a lemma.

Lemma

Any Cauchy sequence is bounded.

Proof.

Let $(x_{n})_{n}$ be a Cauchy sequence. Then $\exists N\in\mathbb{N}$ such that $n\geq N\implies\lvert x_{n}-x_{N}\rvert<1$ (this is equivalent to the definition of Cauchy with $m=N$ ). Then

\lvert x_{n}\rvert=\lvert x_{n}-x_{N}+x_{N}\rvert\leq\lvert x_{n}-x_{N}\rvert+% \lvert x_{N}\rvert\leq 1+\lvert x_{N}\rvert.

(4.49)

Therefore, for

M=\max\{1+\lvert x_{N}\rvert,\lvert x_{1}\rvert,\lvert x_{2}\rvert,\dots,% \lvert x_{N-1}\rvert\},

(4.50)

and thus for all $n\in\mathbb{N}$ we have $\lvert x_{n}\rvert\leq M$ , so $(x_{n})_{n}$ is bounded. ∎

Now suppose $(x_{n})_{n}$ is a Cauchy sequence, then by the lemma above, $(x_{n})_{n}$ is bounded and thus by the Bolzano-Weierstrass Theorem (4.8) there exists a convergent subsequence $(x_{m_{k}})_{k}$ which converges to some limit $x$ . We now want to show that the Cauchy sequence $(x_{n})_{n}$ also converges to $x$ . Let $\varepsilon>0$ , Since $(x_{n})_{n}$ is Cauchy $\exists N\in\mathbb{N}$ such that $m,n\geq N\implies\lvert x_{m}-x_{n}\rvert<\frac{\varepsilon}{2}$ . Also, since $x_{m_{k}}\to x,\ \exists K\in\mathbb{N}$ such that $k\geq K\implies\lvert x_{m_{k}}-x\rvert<\frac{\varepsilon}{2}$ . Then, for $n\geq\max\{N,K\}$ we have

\lvert x_{n}-x\rvert=\lvert x_{n}-x_{m_{n}}+x_{m_{n}}-x\rvert\leq\lvert x_{n}-% x_{m_{k}}\rvert+\lvert x_{m_{k}}-x\rvert<\frac{\varepsilon}{2}+\frac{% \varepsilon}{2}=\varepsilon,

(4.51)

and thus $(x_{n})_{n}$ is convergent with limit $x$ .
(Reverse direction $\impliedby$ ): Suppose $(x_{n})_{n}$ is a convergent sequence with limit $x$ . Let $\varepsilon>0$ , then $\exists N\in\mathbb{N}$ such that $n\geq N\implies\lvert x_{n}-x\rvert<\frac{\varepsilon}{2}$ . For $m,n\geq N$ we have

\lvert x_{m}-x_{n}\rvert=\lvert x_{m}-x+x-x_{n}\rvert\leq\lvert x_{m}-x\rvert+% \lvert x-x_{n}\rvert<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,

(4.52)

and thus $(x_{n})_{n}$ is a Cauchy sequence. ∎