\chapter{Complex Sequences and series}


\section{Complex sequences}

Sequences are first covered in Set Theory as an ordered list of (possibly infinite) values, where repetition is allowed. It is then formally defined as a function mapping integers to elements of an arbitrary set. Set theory being as fundamental as it is, it didn't have much more to say about sequences.

Elementary Number Theory goes further and delegates a chapter to 'integer sequences'; sequences of integers and rational numbers (so much for the 'integer part)'. This book discusses properties of any real sequence and particularly focuses on their behaviour as the amount of terms approaches infinity; elementary number theory does not concern itself with such concepts. This chapter introduces sequences in the space of real numbers and aims to build a correct rigorous understand of precisely what it means for a sequence to 'converge' to a value.

This chapter additionally aims to act as a stepping stone for dealing with real functions. Real sequences are still functions, and despite being simpler, we will see that they form the fundamentals for rigorously defining ideas of analysis such as limits. Even if our sequences terms are all happen  to be rational, it is still interesting to note that the sequence might be approaching an irrational number. These are the kinds of fun things that happen when you get 'real' about mathematics (okay, that was pretty lame).


- real sequence

\begin{definition}[Complex sequence]
A \emph{complex sequence} is a sequence $(z_n)^{\infty}_{n=1}$ of complex numbers, so each $ z_n \in \mathbb{C}$
\end{definition}


The order on the field of reals cannot be extended to the field of complex numbers, so bounded above and below do not make sense in $\mathbb{C}$. Since the complex norm (modulus) is always real, so a bounded complex sequence still has meaning.
\begin{definition}[Complex bounded sequence]
A \emph{complex bounded sequence} is a complex sequence whose modulus is bounded for all terms, equivalently, there is some $M$ such that for all $n$ we have $|x_n| \leq M$.
\end{definition}

The notion of convergent sequences translates perfectly fine.

\begin{definition}[Complex convergent sequence]
For a complex sequence $(a_n)_{n \in \mathbb{N}}$, its \emph{limit} is a number $L$ such that for any positive $\varepsilon$, we can find an integer $N$ so that whenever $n >N$ we have  $| a_n -L| < \varepsilon$. Basically, $L$ is arbitrarily close all remaining terms of a sequence. A \emph{convergent complex sequence} is a complex sequence with a limit.
	\[ \lim_{n \to \infty} a_n = L \iff \forall \varepsilon \in (0,\infty) ( \exists N \in \mathbb{N} [ n > N \implies  |a_n - L| < \varepsilon ]  ]  )  \]
\end{definition}

\begin{definition}[Complex Cauchy sequence]
A \emph{complex Cauchy sequence} is a complex sequence $(a_n)_{n \in \mathbb{N}}$, where for any positive $\varepsilon$, we can find an integer $N$ so that whenever $n,m > N$ we have $|a_n - a_m| < \varepsilon$. Basically, the absolute difference between all remaining terms can be bouned arbitrarily close to $0$ given enough terms of the sequence have passed.
	\[ (x_n)_{n\in \mathbb{N}} \text{ is Cauchy} \iff \forall \varepsilon \in (0,\infty) ( \exists N \in \mathbb{N} [ n,m > N \implies  |a_n - a_m| < \varepsilon ]  ]  )  \]
\end{definition}

\begin{proposition}
Complex convergent sequences are complex Cauchy sequences.
\end{proposition}

We can reduce the calculation of limits of complex sequences into 2 limits of real sequences. Mapping complex analysis problems in terms of real analysis is a common technique that will be employed throughout this book.

\begin{proposition}
\[ \lim_{n \to \infty} c_n = L \iff [ \lim_{n \to \infty} \Re(c_n) = \Re(L) ] \land [ \lim_{n \to \infty} \Im(c_n) = \Im(L) ]  \]
\end{proposition}


\section{Series}

- complex series
- convergent series



- absolutely convergent series
- absolute convergence implies convergence

\subsection{Convergent series tests}

Unfortunately many convergence tests of real analysis fail to hold in $\mathbb{C}$, however we can  change our problem into a problem of 2 real series and prove convergence for those.

\begin{proposition}
	\[  \sum^{\infty}_{n=0} c_n = L \iff [ \sum^{\infty}_{n=0} \Re(c_n) = \Re(L) ] \land [ \sum^{\infty}_{n=0} \Im(c_n) = \Im(L) ]  \]
\end{proposition}

Fortunately the comparison test holds for complex series.

- complex comparison test