Sequences and series

Section 1.5 Sequences and series

The complex modulus function \(\abs{z}\text{,}\) as well as the distance function \(d(z,w)=\abs{z-w}\) it defines, gives rise to theory of convergence of complex sequences and series that is a straightforward generalization of real sequences and series theory.

Subsection Sequences

Definition 1.5.1. Limit of sequence.

A sequence of complex numbers \((z_n)_{n=1}^\infty=z_1,z_2,\dots\) of complex numbers converges if there is a complex number \(z\in \C\) satisfying the following property: for all \(\epsilon > 0\text{,}\) there exists a \(N > 0\) such that if \(n\geq N\text{,}\) then \(\abs{z_n-z}< \epsilon\text{.}\) We write \(\lim\limits_{n\to \infty}z_n=z\) (or \(z_n\to z\)) in this case, and call \(z\) the limit of the sequence.

A sequence diverges if it does not converge.

Definition 1.5.2. Infinite limit.

A complex sequence \((z_n)_{n=1}^\infty\) has an infinite limit, denoted \(\lim\limits_{n\to \infty}z_n=\infty\) if for all \(R> 0\text{,}\) there exists an \(N> 0\) such that if \(n\geq N\text{,}\) then \(\abs{z_n}> R\text{.}\)

Remark 1.5.3. Infinite limit.

Identifying an element \(z\) of \(\C^*=\C\cup \{\infty\}\) with its corresponding point \(P_z\in S^2\subseteq \R^3\) using the stereographic projection, you can show that

\begin{align*} z_n\to \infty \amp \iff P_{z_n}\to N=(0,0,1)\text{.} \end{align*}

Remark 1.5.4. Logical shorthand.

Definition 1.5.1–1.5.2, can be expressed using logical shorthand as follows:

\begin{align} z_n\to z \amp \iff \forall \epsilon > 0\, \exists N > 0( n\geq N \implies \abs{z_n-z}< \epsilon) \tag{1.19}\\ z_n\to\infty \amp \iff \forall R > 0\, \exists N > 0( n\geq N \implies \abs{z_n}> R) \text{.}\tag{1.20} \end{align}

Limits of complex sequences satisfy all the usual familiar limit properties of real sequences, and then some.

Theorem 1.5.5. Limit properties.

Let \((z_n)_{n=1}^\infty\) and \((w_n)_{n=1}^\infty\) be sequences of complex numbers.

\(z_{n}\rightarrow z\) if and only if \(\val{z_n-z}\rightarrow 0\text{.}\)

In particular, \(z_{n}\rightarrow 0\) if and only if \(\val{z_n}\rightarrow 0\text{.}\)
\(\{z_{n}\}\) converges if and only if \(\{\Re z_{n}\}\) and \(\{\Im z_{n}\}\) both converge, in which case

\begin{align*} \lim_{n\rightarrow\infty}z_{n}\amp = (\lim_{n\rightarrow\infty}\Re z_{n})+i\lim\limits_{n\to \infty}\Im z_n \text{.} \end{align*}

In particular, we have \(\lim\limits_{n\to \infty}z_n=z\) if and only if

\begin{align*} \lim\limits_{n\to \infty}\Re z_n \amp =\Re z \amp \lim\limits_{n\to \infty}\Im z_n= \Im z\text{.} \end{align*}
\(z_n\to z\) if and only if \(\overline{z_n}\to \overline{z}\text{.}\)
If \(z_{n}\rightarrow z\text{,}\) then \(\val{z_n}\rightarrow \val{z}\text{.}\)
If \(\lim\limits_{n\to \infty}z_n=z\) and \(\lim\limits_{n\to \infty}w_n=w\text{,}\) then

\begin{align} \lim\limits_{n\to \infty} cz_{n}+\lim\limits_{n\to \infty}dw_{n}\amp = c\lim\limits_{n\to \infty}z_n+d\lim\limits_{n\to \infty}w_n=cz+dw \text{ for any } c,d\in \C\tag{1.21}\\ \lim\limits_{n\to \infty}z_{n}w_{n}\amp = (\lim\limits_{n\to \infty}z_n)(\lim\limits_{n\to \infty}w_m)=zw\amp \tag{1.22}\\ \lim\limits_{n\to \infty}\frac{z_n}{w_n}\amp = \frac{\lim\limits_{n\to \infty}z_n}{\lim\limits_{n\to \infty}w_n}= z/w, \text{ assuming } w\ne 0 \text{.}\tag{1.23} \end{align}
\(z_{n}\rightarrow \infty\) if and only if \(\frac{1}{z_{n}}\rightarrow 0\text{,}\) assuming \(z_n\ne 0\) for all \(n\text{.}\)

Proof.

Most of these statements are proved exactly as with their analogous statements for real sequences, and both in turn use little beyond properties of \(\abs{\ }\text{,}\) either as the absolute value (on the reals) or its extension, the complex modulus. We content ourselves with proving (2) by way of illustration.

For the reverse direction (\(\Leftarrow\)), if both \(\lim\limits_{n\to \infty}\Re z_n\) and \(\lim\limits_{n\to \infty}\Im z_n\) exist, then we have

\begin{align*} \lim\limits_{n\to \infty}z_n\amp = \lim\limits_{n\to \infty}\Re z_n+i\Im z_n \amp \\ \amp =\lim\limits_{n\to \infty}\Re z_n+i\lim\limits_{n\to \infty}\Im z_n \amp (\knowl{./knowl/xref/eq_lim_seq_linear.html}{\text{(1.21)}}) \text{,} \end{align*}

as desired.

Consider now the forward implication (\(\Rightarrow\)). We assume \(\lim\limits_{n\to \infty}z_n=z\) and wish to show that \(\Re z=\lim\limits_{n\to \infty}\Re z_n\) and \(\Im z=\lim\limits_{n\to \infty}\Im z_n\) (which also proves these limits exists). We will give a proof of the second equality: you can prove the first in much the same way. Fix any positive \(\epsilon > 0\text{.}\) Since \(z_n\rightarrow z\text{,}\) we can find an \(N> 0\) such that \(\abs{z-z_n}< \epsilon\) for all \(n\geq N\text{.}\) But then we have

\begin{align*} \abs{\Im z-\Im z_n} \amp =\abs{\Im(z-z_n)}\\ \amp \leq \abs{z-z_n} \\ \amp < \epsilon \end{align*}

for all \(n\geq N\text{,}\) showing that \(\Im z_n\to \Im z\text{,}\) as desired.

Example 1.5.6. Limit of sequence.

Define \(z_n=\frac{1}{2^n}(1+i)^n\text{.}\) Show that \((z_n)_{n=1}^{\infty}\) converges and compute its limit.

Solution.

We have \(z_n=w^n\text{,}\) where \(w=\frac{1}{2}(1+i)=\frac{\sqrt{2}}{2}(\cos(\pi/4)+i\sin(\pi/4))\text{.}\) It follows from de Moivre’s formula that

\begin{equation*} z_n=w^n=(\sqrt{2}/2)^n(\cos(n\pi/4)+i\sin(n\pi/4)) \end{equation*}

and hence that

\begin{equation*} \abs{z_n}=(\sqrt{2}/2)^n. \end{equation*}

Since \((\sqrt{2}/2)< 1\text{,}\) we have \(\abs{z_n}\to 0\text{.}\) It follows from Theorem 1.5.5 that \(z_n\to 0\text{.}\)

Example 1.5.7. Divergent sequence.

Show that the sequence \((i^n)_{n=1}^{\infty}\) diverges.

Solution.

Observe first that the sequence is periodic of the form \[ i,-1,-i,1,i,-1,\dots \] Assume by contradiction that \(i^n\to w\) for some complex number \(w\text{.}\) Taking \(\epsilon =1\) in the definition of sequence convergence, this would imply the existence of an \(N\) such that \(\abs{i^n-w}< 1\) for all \(n\geq N\text{.}\) In particular, we would have \(\abs{1-w}< 1\) and \(\abs{-1-w}< 1\text{.}\) But this would imply that

\begin{align*} 2=\abs{1+1} \amp = \abs{1-w-(-1-w)} \\ \amp\leq \abs{1-w}+\abs{-1-w} \\ \amp < 1+1=2\text{,} \end{align*}

a contradiction since then we’d have \(2< 2\text{.}\) We conclude that there is no limit to the sequence.

Example 1.5.8. Sequence with infinite limit.

Show that \(\lim\limits_{n\to \infty}\frac{\sqrt{n}i+n^2}{5n+i}=\infty\text{.}\)

Solution.

Let \(z_n=\frac{\sqrt{n}i+n^2}{5n+i}\text{.}\) We will show that \(\frac{1}{z_n}\to 0\text{,}\) and thus \(z_n\to \infty\text{,}\) by Theorem 1.5.5. Looking at

\begin{equation*} \frac{1}{z_n}=\frac{5n+i}{\sqrt{n}i+n^2} \end{equation*}

we see that the dominating term on top is the \(5n\) and the dominating term on the bottom is \(n^2\text{.}\) We will make this argument rigorous by factoring out these dominating terms from the top and bottom and computing the limit using a combination of algebra and limit rules.

\begin{align*} \lim\limits_{n\to \infty}\frac{1}{z_n}\amp= \lim\limits_{n\to \infty}\frac{5n+i}{\sqrt{n}i+n^2} \\ \amp = \lim\limits_{n\to \infty}\frac{n\left(5+\frac{i}{n}\right)}{n^2\left(\frac{i}{n^{3/2}}+1\right)} \\ \amp = \lim\limits_{n\to \infty}\frac{1}{n} \cdot \frac{5+\frac{i}{n}}{\frac{i}{n^{3/2}}+1}\\ \amp = \lim\limits_{n\to \infty}\frac{1}{n} \cdot \lim\limits_{n\to \infty}\frac{5+\frac{i}{n}}{\frac{i}{n^{3/2}}+1} \amp (\text{prod. rule})\\ \amp =\lim\limits_{n\to \infty}\frac{1}{n}\cdot \frac{5+i\lim\limits_{n\to \infty}\frac{1}{n}}{i\lim\limits_{n\to \infty}\frac{1}{n^{3/2}}+1} \amp (\text{quot., sum, scal. rule}) \amp (n\to 0) \\ \amp =0\cdot \frac{5}{1}\\ \amp =0 \text{.} \end{align*}

Since \(1/z_n\to 0\text{,}\) we conclude that \(z_n\to \infty\text{.}\)

Subsection Series

As with the reals, from a theory of convergent sequences we easily derive a theory of convergent series. The definitions and results below are all straightforward generalizations of facts about series of real numbers.

Definition 1.5.9. Series.

Given a complex sequence \((z_n)_{n=1}^{\infty}\text{,}\) the formal expression \(\sum_{n=1}^\infty z_n\) is called a (complex) series.

For each \(k\geq 1\) we define the \(k\)-th partial sum \(S_k\) of the series \(\sum_{n=1}^\infty z_n\) as

\begin{equation*} S_k=\sum_{n=1}^k z_k\text{.} \end{equation*}

We say the series converges if

\begin{equation*} \lim\limits_{k\to \infty}S_k=\lim\limits_{k\to \infty}\sum_{n=1}^k z_k=z \end{equation*}

for some complex number \(z\in \C\text{,}\) in which case we write

\begin{equation*} \sum_{n=1}^\infty z_n=z \end{equation*}

and say that the series converges to \(z\text{.}\)

A series diverges if it does not converge.

Theorem 1.5.10. Series properties.

Let \((z_n)_{n=1}^\infty\) and \((w_n)_{n=1}^\infty\) be complex sequences.

\(\sum_{n=1}^\infty z_n=z\) if and only if \(\lim\limits_{k\to \infty}(S_k-z)=0\text{,}\) where \(S_k\) is the \(k\)-th partial sum of \(\sum_{n=1}^\infty z_n\text{.}\)
\(\sum_{n=1}^\infty z_n\) converges if and only if \(\sum_{n=1}^\infty \Re z_n\) and \(\sum_{n=1}^\infty \Im z_n\) converge, in which case

\begin{equation*} \sum_{n=1}^\infty z_n=\sum_{n=1}^\infty \Re z_n +i \sum_{n=1}^\infty \Im z_n\text{.} \end{equation*}

In particular, \(\sum_{n=1}^\infty z_n=z\) if and only if

\begin{align*} \sum_{n=1}^\infty \Re z_n \amp = \Re z \amp \sum_{n=1}^\infty \Im z_n=\Im z \end{align*}
If \(\sum_{n=1}^\infty z_n=z\) and \(\sum_{n=1}^\infty w_n=w\text{,}\) then

\begin{align*} \sum_{n=1}^\infty cz_n+\sum_{n=1}^\infty dw_n \amp = c\sum_{n=1}^\infty z_n+d\sum_{n=1}^\infty w_n \\ \amp =cz+dw \text{,} \end{align*}

for any \(c,w\in \C\text{.}\)
Divergence test.
If \(\sum_{n=1}^\infty z_n\) converges, then \(\lim\limits_{n\to \infty}z_n=0\text{.}\)
Absolute convergence test.

If \(\sum_{n=1}^\infty \abs{z_n}\) converges, then \(\sum_{n=1}^\infty z_n\) converges and

\begin{equation*} \abs{\sum_{n=1}^\infty z_n}\leq \sum_{n=1}^\infty \abs{z_n}\text{.} \end{equation*}

Proof.

Example 1.5.11. Geometric series.

Let \(z\in \C\text{.}\) Show that the geometric series \(\sum_{n=0}^\infty z^n\) converges to \(\frac{1}{1-z}\) if \(\abs{z}< 1\text{,}\) and diverges for \(\abs{z}\geq 1\text{.}\)

Solution.

First observe that if \(\abs{z}\geq 1\text{,}\) then the sequence \((\abs{z^n})\not\to 0\text{.}\) It follows that \(z_n\not\to 0\) and thus that the series diverges by the divergence test.

Now assume \(\abs{z}< 1\text{.}\) For all \(k\geq 1\) the \(k\)-th partia sum is

\begin{equation*} S_k=1+z+z^2+\cdots + z^k=\frac{z^{k+1}-1}{z-1}, \end{equation*}

using the polynomial identity (1.18). It follows that

\begin{align*} \sum_{n=0}^\infty z^n \amp =\lim_{k\to\infty}S_k\\ \amp =\lim_{k\to\infty}\frac{z^{k+1}-1}{z-1}\\ \amp =\frac{\lim_{k\to\infty}z^{k+1}-1}{\lim_{k\to\infty}z-1}\\ \amp =\frac{0-1}{z-1} \amp (\abs{z}< 0\implies z^{k+1}\to 0)\\ \amp =\frac{1}{1-z}\text{.} \end{align*}

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