Sequences and Series of Complex Numbers

The concept of a sequence or series of real numbers can easily be extended to complex numbers. Many of the definitions and important theorems also carry over. It will be assumed that many of these notions are already understood so the material will only be glossed over. The Real Analysis section on this site provides a more in-depth look into these topics.

Definition: A Sequence of Complex Numbers is an infinite ordered list of complex numbers, $(a_n)_{n=1}^{\infty} = (a_1, a_2, ..., a_n, ...)$, $a_n \in \mathbb{C}$ for all $n \in \mathbb{N}$.

The start index for a sequence is conventionally $1$, but it is perfectly fine to consider sequences of complex numbers whose start index is $0$ or any other integer.

For example, the sequence $\displaystyle{\left ( \frac{n + i}{n - i} \right )_{n=1}^{\infty}}$ when expanded is:

(1)

\begin{align} \quad \left ( \frac{n + i}{n - i} \right )_{n=1}^{\infty} = \left ( \frac{1 + i}{1 - i}, \frac{2 + i}{2 - i} , ..., \frac{n + i}{n - i}, ... \right ) \end{align}

Definition: A sequence of complex numbers $(a_n)_{n=1}^{\infty}$ is said to Converge to $A \in \mathbb{C}$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid a_n - A \mid < \epsilon$. A sequence of complex numbers $(a_n)_{n=1}^{\infty}$ is said to Diverge if it does not converge to any $A \in \mathbb{C}$.

For example, consider the sequence $\displaystyle{\left ( \frac{(2 + i)}{n+1} \right )_{n=1}^{\infty}}$. We claim that this series converges to $2 + i$. To prove this, let $\epsilon > 0$ be given. Then:

(2)

\begin{align} \quad \biggr \lvert \frac{(2 + i)n}{n+1} - (2 + i) \biggr \rvert &= \biggr \lvert \frac{(2 + i)n}{n+1} - \frac{(2 + i)(n+1)}{n+1} \biggr \rvert \\ &= \biggr \lvert \frac{(2 + i)n - (2 + i)(n + 1)}{n + 1} \biggr \rvert \\ &= \biggr \lvert \frac{-(2 + i)}{n + 1} \biggr \rvert \\ &= \frac{\sqrt{5}}{n + 1} \end{align}

So choose $N \in \mathbb{N}$ such that $N > \frac{\sqrt{5}}{\epsilon}$. Then $\frac{1}{N} < \frac{\epsilon}{\sqrt{5}}$. So if $n \geq N$ then $n + 1 \geq N$ and so $\displaystyle{1}{n+1} < \frac{\epsilon}{\sqrt{5}}$ and so:

(3)

\begin{align} \quad \biggr \lvert \frac{(2 + i)n}{n+1} - (2 + i) \biggr \rvert < \sqrt{5} \cdot \frac{\epsilon}{\sqrt{5}} = \epsilon \end{align}

Therefore $\displaystyle{\left ( \frac{(2 + i)n}{n+1} \right )_{n=1}^{\infty}}$ converges to $2 + i$.

Definition: If $(a_n)_{n=1}^{\infty}$ is a sequence of complex numbers then the corresponding Series of Complex Numbers is the infinite formal sum $\displaystyle{\sum_{n=1}^{\infty} a_n}$. The corresponding Sequence of Partial Sums is the sequence of complex numbers $(s_n)_{n=1}^{\infty}$ where $\displaystyle{s_n = \sum_{k=1}^{n} a_n}$ for each $n \in \{ 1, 2, ... \}$. The complex series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to Converge to the sum $S \in \mathbb{C}$ if the sequence of partial of sums $(s_n)_{n=1}^{\infty}$ converges to $S$, and if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to Diverge if it does not converge to any $S \in \mathbb{C}$. The series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to be Absolutely Convergent if $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges, and Conditionally Convergent if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges and $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ diverges.