Cauchy Sequences of Real Numbers

# Cauchy Sequences of Real Numbers

One very important classification of sequences are known as Cauchy Sequences which we defined as follos:

 Definition: A sequence $(a_n)$ is called a Cauchy Sequence if $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\forall m, n ≥ N$, then $\mid a_n - a_m \mid < \epsilon$. A sequence is not Cauchy if $\exists \epsilon_0 > 0$ such that $\forall N \in \mathbb{N}$ there exists at least one $m$ and one $n$ where $m, n > N$ such that $\mid a_n - a_m \mid ≥ \epsilon_0$.

As you might suspect, if $(a_n)$ and $(b_n)$ are Cauchy sequences, then the sequences $(a_n + b_n)$, $(a_n - b_n)$, $(ka_n)$ and $(a_nb_n)$ are also Cauchy. The proofs of these can be found on the Additional Cauchy Sequence Proofs page.

We will now look at some more important lemmas about Cauchy sequences that will lead us to the The Cauchy Convergence Criterion.

 Lemma 1: Every convergent sequence $(a_n)$ of real numbers is also a Cauchy sequence.
• Proof: Let $(a_n)$ be a convergent sequence to the real number $A$. Then $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - A \mid < \epsilon$. So, for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - A \mid < \epsilon_1 = \frac{\epsilon}{2}$ and so if $m, n ≥ N$ then:
(1)
\begin{align} \quad \mid a_n - a_m \mid = \mid a_n - A + A - a_m \mid ≤ \mid a_n - A \mid + \mid A - a_m \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore the convergent sequence $(a_n)$ is also a Cauchy sequence. $\blacksquare$
 Lemma 2: If $(a_n)$ is a Cauchy sequence of real numbers then $(a_n)$ is also bounded.
• Proof: Suppose that $(a_n)$ is a Cauchy sequence. Then $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid a_n - a_m \mid < \epsilon$.
• Choose $\epsilon = 1$, and so there exists an $N \in \mathbb{N}$ such that if $n, N ≥ N$ then $\mid a_n - a_N \mid < 1$. Using the reverse triangle inequality we have that $\mid a_n \mid - \mid a_N \mid < 1$, and so $\mid a_n \mid < 1 + \mid a_N \mid$ for all $n ≥ N$.
• Now let $M = \mathrm{max} \{ \mid a_1 \mid, \mid a_2 \mid, ..., \mid a_{N-1}\mid, 1 + \mid a_N \mid \}$. Therefore $\mid a_n \mid ≤ M$ for all $n \in \mathbb{N}$ and so $(a_n)$ is a bounded sequence. $\blacksquare$

Before we look at the The Cauchy Convergence Criterion, let's first take a step back and look at some examples of Cauchy sequences and non-Cauchy sequences:

## Example 1

Show that the sequence $\left ( \frac{1}{n} \right )$ is a Cauchy sequence.

We want to show that $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\forall m, n ≥ N$, then $\mid x_n - x_m \mid < \epsilon$.

Let $\epsilon > 0$ be given, and choose $N$ such that $N > \frac{2}{\epsilon}$. If $m, n ≥ N$ then $n ≥ N > \frac{2}{\epsilon}$ implies that $\frac{1}{n} ≤ \frac{1}{N} < \frac{\epsilon}{2}$ and similarly $m ≥ N > \frac{2}{\epsilon}$ implies that $\frac{1}{m} ≤ \frac{1}{N} < \frac{\epsilon}{2}$. Therefore applying the triangle inequality we have that

(2)
\begin{align} \quad \biggr \rvert x_n - x_m \biggr \rvert = \biggr \rvert \frac{1}{n} - \frac{1}{m} \biggr \rvert ≤ \biggr \rvert \frac{1}{n} \biggr \rvert + \biggr \rvert \frac{1}{m} \biggr \rvert = \frac{1}{n} + \frac{1}{m} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

Therefore $\left ( \frac{1}{n} \right )$ is a Cauchy sequence.

## Example 2

Show that the sequence $((-1)^n)$ is not Cauchy.

Notice that if $n$ is even then $a_n = 1$, and so $a_{n+1} = -1$. Choose $\epsilon_0 = 2$, and select any even $n$ such that $n ≥ N \in \mathbb{N}$. Then choose $m = n + 1$. Therefore $\mid x_n - x_m \mid = \mid 1 - (-1) \mid = 2 ≥ \epsilon_0 = 2$. So $((-1)^n)$ is not Cauchy.

## Example 3

Show that the sequence $\left ( \frac{1}{n^2} \right )$ is a Cauchy sequence.

We want to show that $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\forall m, n ≥ N$, then $\mid x_n - x_m \mid < \epsilon$.

Let $\epsilon > 0$ be given and choose $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\biggr \rvert \frac{1}{n^2} - \frac{1}{m^2} \biggr \rvert < \epsilon$.

Choose $N > \frac{2}{\epsilon}$. Therefore if $m, n ≥ N$ then $n ≥ N > \frac{2}{\epsilon}$ implies that $\frac{1}{n} ≤ \frac{1}{N} < \frac{\epsilon}{2}$ and $m ≥ N > \frac{2}{\epsilon}$ implies that $\frac{1}{m} ≤ \frac{1}{N} < \frac{\epsilon}{2}$ and so for $m, n ≥ N$ we have:

(3)
\begin{align} \quad \biggr \rvert \frac{1}{n^2} - \frac{1}{m^2} \biggr \rvert ≤ \biggr \rvert \frac{1}{n^2} \biggr \rvert + \biggr \rvert \frac{1}{m^2} \biggr \rvert = \frac{1}{n^2} + \frac{1}{m^2} ≤ \frac{1}{n} + \frac{1}{m} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore $\left ( \frac{1}{n} \right )$ is a Cauchy sequence.

## Example 4

Give an example to show that the converse of lemma 2 is false.

The converse of lemma 2 says that "if $(a_n)$ is a bounded sequence, then $(a_n)$ is a Cauchy sequence of real numbers."

The sequence $((-1)^n)$ provided in Example 2 is bounded and not Cauchy.