The Cauchy Convergence Criterion

# The Cauchy Convergence Criterion

Recall that a sequence of real numbers $(a_n)$ is said to be a **Cauchy Sequence** if $\forall \epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n ≥ N$ then $\mid a_n - a_m \mid < \epsilon$. On the Cauchy Sequences page, we already noted that every convergent sequence of real numbers is Cauchy, and that every Cauchy sequence of real numbers is bounded. We will now look at another important theorem known as the **Cauchy Convergence Criterion**.

Theorem (Cauchy Convergence Criterion): If $(a_n)$ is a sequence of real numbers, then $(a_n)$ is convergent if and only if $(a_n)$ is a Cauchy sequence. |

Note that the Cauchy Convergence Criterion will allow us to determine whether a sequence of real numbers is convergent whether or not we have a suspected limit in mind for a sequence.

**Proof:**$\Rightarrow$ Suppose that $(a_n)$ is a convergent sequence of real numbers. On the Cauchy Sequences page, we already verified that a convergent sequence of real numbers is Cauchy.

- $\Leftarrow$ Suppose that $(a_n)$ is a Cauchy sequence. We want to show that $(a_n)$ is thus convergent to some real number in $\mathbb{R}$. Now since $(a_n)$ is a Cauchy sequence it follows that $(a_n)$ is bounded. Since $(a_n)$ is bounded, it follows from The Bolzano-Weierstrass Theorem that there exists a subsequence of $(a_n)$, call it $(a_{n_k})$ that converges to some real number $A^*$.

- Since $(a_n)$ 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 $\epsilon_1 = \frac{\epsilon}{2} > 0$, and so there exists an $N_1 \in \mathbb{N}$ such that if $m, n ≥ N_1$ then $\mid a_n - a_m \mid < \epsilon_1 = \frac{\epsilon}{2}$.

- Now look at the subsequence $(a_{n_k})$ which converges to $A^*$. There thus exists a natural number $K ≥ N_1$ where $K$ belongs to the set of indices $\{ n_1, n_2, ... \}$ such that $\mid a_{K} - A^* \mid < \frac{\epsilon}{2}$.

- Since $K ≥ N_1$ then if we substitute $m = K$ we have that for $n ≥ N_1$:

\begin{align} \mid a_n - a_K \mid < \frac{\epsilon}{2} \end{align}

- And so for $n ≥ N_1$ we have that:

\begin{align} \quad \mid a_n - A^* \mid = \mid a_n - a_K + a_K - A^* \mid ≤ \mid a_n - a_K \mid + \mid a_K - A^* \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- Therefore $\lim_{n \to \infty} a_n = A^*$, and so $(a_n)$ is convergent to the real number $A^*$. $\blacksquare$

We will summarize the lemma from the Cauchy Sequences page and the Cauchy Convergent Criterion as follows:

- Any sequence of real numbers $(a_n)$ that is convergent is also a Cauchy sequence.

- Any Cauchy sequence of real numbers $(a_n)$ is also a convergent sequence.