# Series

## Convergence of sequences

An infinite sequence $\left\{z_1,z_2,z_3 \ldots\right\}$ of complex numbers has a limit $z$ if, for each positive number $\varepsilon$, there exists a positive integer $n_0$ such that \begin{eqnarray}\label{seq} \left|z_n-z\right|< \varepsilon \quad \text{whenever}\quad n > n_0. \end{eqnarray}

Geometrically, this means that for sufficiently large values of $n$, the points $z_n$ lie in any given $\varepsilon$ neighborhood of $z$ (Figure 1). Since we can choose $\varepsilon$ as small as we please, it follows that the points $z_n$ become arbitrarily close to $z$ as their subscripts increase. Note that the value of $n_0$ that is needed will, in general, depend on the value of $\varepsilon$.

The sequence $\left\{z_n\right\}_{n=1}^{\infty}$ can have at most one limit. That is, a limit $z$ is
unique
if it exists. When that limit exists, the sequence is said to *converge* to $z$; and we write
\begin{eqnarray*}
\lim_{n\rightarrow \infty} z_n=z
\end{eqnarray*}
If the sequence has no limit, it *diverges*.

**Proof**

To prove this theorem, we first assume that conditions (\ref{teoseq02}) hold. That is, there exist, for each $\varepsilon>0$, positive integers $n_1$ and $n_2$ such that \[ |x_n-x|<\frac{\varepsilon}{2}\quad \text{whenever} \quad n>n_1 \] and \[ |y_n-y|<\frac{\varepsilon}{2}\quad \text{whenever} \quad n>n_2. \] Hence if $n_0$ is the larger of the two integers $n_1$ and $n_2$, \[ |x_n-x|<\frac{\varepsilon}{2}\quad \text{and} \quad |y_n-y|<\frac{\varepsilon}{2} \quad \text{whenever} \quad n > n_0. \] Since \[ |(x_n+iy_n)-(x+iy)|=|(x_n-x)+(y_n-y)|\leq |x_n-x|+|y_n-y|, \] then \[ |z_n-z|< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \quad \text{whenever} \quad n > n_0. \] Therefore condition (\ref{teoseq01}) holds.

Conversely, if we start with condition (\ref{teoseq01}), we know that for each positive number $\varepsilon$, there exists a positive integer $n_0$ such that \[ |(x_n+iy_n)-(x+iy)|<\varepsilon \quad \text{whenever} \quad n>n_0. \] However \[ |x_n-x|\leq |(x_n-x)+(y_n-y)|=|(x_n+iy_n)-(x+iy)| \] and \[ |y_n-y|\leq |(x_n-x)+(y_n-y)|=|(x_n+iy_n)-(x+iy)|. \] Consequently \[ |x_n-x|<\varepsilon \quad \text{and} \quad |y_n-y|<\varepsilon \quad \text{whenever} \quad n > n_0. \] Therefore, conditions (\ref{teoseq02}) are satisfied. $\blacksquare$

## Convergence of series

An infinite series \begin{eqnarray}\label{series01} \sum_{n=1}^{\infty}z_n=z_1+z_2+z_3+\cdots \end{eqnarray} of complex numbers converges to the sum $S$ if the sequence

**Proof**

To prove this theorem, we first write the partial sums (\ref{partialsum}) as \begin{eqnarray}\label{teo03} S_N = X_N+iY_N, \end{eqnarray} where \[ X_N = \sum_{n=1}^{N}x_n \quad \text{and}\quad Y_N = \sum_{n=1}^{N}y_n. \] Now statement (\ref{teo01}) is true if and only if \begin{eqnarray}\label{teo04} \lim_{N\rightarrow \infty} S_N = S; \end{eqnarray} and, in view of relation (\ref{teo03}) and Theorem 1 on sequences, limit (\ref{teo04}) holds if and only if \begin{eqnarray}\label{teo05} \lim_{N\rightarrow \infty} X_N=X\quad \text{and}\quad \lim_{N\rightarrow \infty} Y_N=Y. \end{eqnarray} Limits (\ref{teo05}) therefore imply statement (\ref{teo01}), and conversely. Since $X_N=X$ and $Y_N=Y$ are partial sums of the series (\ref{teo02}), the theorem is proved. $\blacksquare$

This theorem can be useful in showing that a number of familiar properties of series in calculus carry over to series whose terms are complex numbers.

**Property 1:** If a series of complex numbers converges, the $n$-th term converges to
zero as
$n$ tends to infinity.

**Property 1**that the terms of convergent series are bounded. That is, when series (\ref{series01}) converges, there exists a positive constant $M$ such that $$|z_n| \leq M \; \text{ for each positive integer } n.$$

Another important property of series of complex numbers that follows from a corresponding property in calculus is the following.

**Property 2:** The absolute convergence of a series of complex numbers implies the
convergence
of that series.

Recall that series (\ref{series01}) is said to be *absolutely convergent* if the series
\begin{eqnarray*}\label{series02}
\sum_{n=1}^{\infty}|z_n|=\sum_{n=1}^{\infty}\sqrt{x^2_n+y^2_n}\quad \quad(z_n=x_n+iy_n)
\end{eqnarray*}
of real numbers $\sqrt{x^2_n+y^2_n}$ converges.

To establish the fact that the sum of a series is a given number $S$, it is often convenient to
define the
*remainder* $\rho_N$ after $N$ terms, using the partial sums:
\begin{eqnarray*}\label{series03}
\rho_N=S-S_N
\end{eqnarray*}
Thus $S=S_N+\rho_N$. Now, since $|S_N-S|=|\rho_N-0|$, then *a series converges to a number $S$ if
and
only if the sequence of remainders tends to zero*.

**Example:** With the aid of remainders, it is easy to verify that
\begin{eqnarray*}
\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}\quad\text{whenever}\quad |z|< 1
\end{eqnarray*}
We need only recall the identity
$$1+z+z^2+\cdots+z^n=\frac{1-z^{n+1}}{1-z}$$
to write the partial sums

## Geometric series exploration

The series introduced in the previous example
\begin{eqnarray*}
\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}\quad\text{whenever}\quad |z|< 1
\end{eqnarray*}
is known as the *geometric series*.

Use the following applet to explore this series. Drag the point $z$ around. Observe what happens when it is inside, outside or on the border of the unit circle. Drag the slider to show the partial sum.

**Code**

Enter the following script in GeoGebra
to explore it yourself and make your own version.
The symbol `#`

indicates comments.

```
#Complex number
Z = 0.72 + ί * 0.61
#Circle of radius 1
c = Circle((0,0), 1)
#Number of terms of the partial series
n = Slider(0, 250, 1, 1, 150, false, true, false, false)
SetValue(n, 250)
#Define the sequence z^n
S = Join({0 + ί * 0, 1 + ί * 0}, Sequence(Z^j, j, 1, n))
#Define partial sum
SP = Sequence(Sum(S, j), j, 1, n + 2)
#Finally join the points of the partial sum
L = Sequence(Segment(Element(SP, j), Element(SP, j + 1)), j, 1, n + 1)
```