Convergence Criterion for Cauchy Sequences of Complex Numbers

Recall from the Cauchy Sequences of Complex Numbers page that a sequence of complex numbers $(z_n)$ is said to be Cauchy if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

(1)

\begin{align} \quad |z_m - z_n| < \epsilon \end{align}

The following proposition is yet another analogue of that for sequences of real numbers. It states that every convergent sequence of complex numbers is Cauchy.

Proposition 1: Let $(z_n)$ be a convergent sequence of complex numbers. Then $(z_n)$ is Cauchy.

• Proof: Since $(z_n)$ converges to some $Z \in \mathbb{C}$ we have that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $|z_n - Z| < \epsilon/2$.

• Therefore, if $m, n \geq N$ we have by the triangle inequality that:

(2)

\begin{align} \quad |z_m - z_n| = |z_m - Z + Z - z_n| \leq |z_m - Z| + |Z - z_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

• So $(z_n)$ is Cauchy. $\blacksquare$

We will see later that the set of complex numbers is special (like $\mathbb{R}$) in that the converse of Proposition 1 holds too, that is, every Cauchy sequence of complex numbers also converges.