Bounded Sequences of Complex Numbers

Bounded Sequences of Complex Numbers

Definition: A sequence of complex numbers $(z_n)$ is said to be Bounded if there exists an $M \in \mathbb{R}$, $M > 0$ such that $|z_n| \leq M$ for all $n \in \mathbb{N}$.

The following proposition tells us that every convergent sequences of complex numbers is bounded.

Proposition 1: Let $(z_n)$ be a sequence of complex numbers that converges. Then $(z_n)$ is bounded.
  • Proof: Let $(z_n)$ converge to $Z \in \mathbb{C}$. Then for $\epsilon = 1 > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
\begin{align} \quad |z_n - Z| < \epsilon = 1 \end{align}
  • By the reversed triangle inequality we see that for all $n \geq N$ that $||z_n| - |Z|| < 1$. So for all $n \geq N$ we have that:
\begin{align} \quad |z_n| \leq 1 + |Z| \end{align}
  • Let:
\begin{align} \quad M = \max \{ |z_1|, |z_2|, ..., |z_{N-1}|, 1 + |Z| \} \end{align}
  • Then $|z_n| \leq M$ for all $n \in \mathbb{N}$, showing that $(z_n)$ is bounded. $\blacksquare$

It is important to note that while every convergent sequence of complex numbers is bounded - the converse is not true in general, i.e., there exists bounded sequences of complex numbers that do not converge. For example, the sequence $(i, -i, i, -i, ...)$ does not converge but is bounded.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License