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Complex Analysis an Open Source Textbook

Russell W. Howell, John H. Mathews

Section 4.3 Geometric Series and Convergence Theorems

We begin this section by presenting a series of the form \(\sum\limits_{n=0}^{\infty}z^n\text{,}\) which is called a geometric series and is one of the most important series in mathematics.
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Proof.

Suppose that \(|z|\lt 1\text{.}\) By Definition 4.1.11, we must show \(\lim\limits_{n \to \infty}S_n=\frac{1}{1-z}\text{,}\) where
\begin{equation} S_n=1+z+z^2+\cdots +z^{n-1}\text{.}\tag{4.3.2} \end{equation}
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Multiplying both sides of Equation (4.3.2) by \(z\) gives
\begin{equation} zS_n = z+z^2+z^3+\cdots+z^{n-1}+z^n\text{.}\tag{4.3.3} \end{equation}
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Subtracting Equation (4.3.3) from Equation (4.3.2) yields
\begin{equation*} (1-z) S_n=1-z^n \end{equation*}
so that
\begin{equation} S_n=\frac{1}{1-z}-\frac{z^n}{1-z}\text{.}\tag{4.3.4} \end{equation}
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Since \(|z|\lt 1\text{,}\) \(\lim\limits_{n \to \infty}z^n=0\text{.}\) (Can you prove this assertion? We ask you to do so in the exercises!) Hence \(\lim\limits_{n \to \infty}S_n=\frac{1}{1-z}\text{.}\)
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Now suppose \(|z| \ge 1\text{.}\) Clearly, \(\lim\limits_{n \to \infty}|z^n| \ne 0\text{,}\) so \(\lim\limits_{n \to \infty}z^n\ne 0\) (see Exercise 4.1.17, Section 4.1). Thus, by the contrapositive of Theorem 4.1.17, \(\sum\limits_{n=0}^{\infty}z^n\) must diverge.
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Proof.

If we let \(\frac{1}{z}\) take the role of \(z\) in Equation (4.3.1), we get
\begin{equation*} \sum_{n=0}^{\infty}\left(\frac{1}{z}\right)^{\!n} = \frac{1}{1-\frac{1}{z}}, \qquad \text{ if } \left|\frac{1}{z}\right| \lt 1\text{.} \end{equation*}
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Multiplying both sides of this equation by \(\frac{1}{z}\) gives
\begin{equation*} \frac{1}{z}\sum_{n=0}^{\infty}\left(\frac{1}{z}\right)^{\!n} = \frac{1}{z-1}, \text{ if } \left|\frac{1}{z}\right|\lt 1\text{.} \end{equation*}
which, by Equation (4.1.10), is the same as
\begin{equation*} \sum_{n=0}^{\infty}\left(\frac{1}{z}\right)^{\! n+1}=\frac{1}{z-1} \text{ if } \left|\frac{1}{z}\right| \lt 1\text{.} \end{equation*}
But this expression is equivalent to saying that \(\sum\limits_{n=1}^{\infty}(\frac{1}{z})^n=\frac{1}{z-1}\text{,}\) if \(1\lt |z|\text{,}\) which is what the corollary claims.
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It is left as an exercise to show that the series diverges if \(|z| \le 1\text{.}\)
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Proof.

Example 4.3.4.

Show that \(\sum\limits_{n=0}^{\infty} \frac{(1-i)^n}{2^n}=1-i\text{.}\)
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Solution.
If we set \(z=\frac{1-i}{2}\text{,}\) then \(|z|=\frac{\sqrt{2}}{2}\lt 1\text{.}\) By Theorem 4.3.1, the sum is
\begin{equation*} \frac{1}{1-\frac{1-i}{2}}=\frac{2}{2-1+i}=\frac{2}{1+i}=1-i \end{equation*}
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Example 4.3.5.

Evaluate \(\sum\limits_{n=3}^{\infty}(\frac{i}{2})^n\text{.}\)
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Solution.
We can put this expression in the form of a geometric series:
\begin{align*} {3} \sum_{n=3}^{\infty}\left(\frac{i}{2}\right)^{\!n} \amp = \sum_{n=3}^{\infty}\left(\frac{i}{2}\right)^{\!3}\left(\frac{i}{2}\right)^{\!n-3} \amp \amp\\ \amp = \left(\frac{i}{2}\right)^{\!3} \sum_{n=3}^{\infty}\left(\frac{i}{2}\right)^{\!n-3} \amp \amp \text{ (by Equation (4.1.10) in \knowl{./knowl/xref/Conv_terms_go_to_0.html}{\text{Theorem 4.1.17}}) }\\ \amp = \left(\frac{i}{2}\right)^{\!3} \sum_{n=0}^{\infty}\left(\frac{i}{2}\right)^{\!n} \amp \amp \text{ (by reindexing) }\\ \amp = \left(\frac{i}{2}\right)^{\!3} \left(\frac{1}{1-\frac{i}{2}}\right) \amp \amp \text{ (by \knowl{./knowl/xref/Geometric_Series.html}{\text{Theorem 4.3.1}} because } \left|\frac{i}{2}\right|=\frac{1}{2}\lt 1)\\ \amp = \frac{1}{20}-\frac{i}{10} \amp \amp \text{ (by standard simplification procedures) } \text{.} \end{align*}
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Remark 4.3.6.

The equalities given in Example 4.3.5 collectively illustrate an important point with regard to evaluating a geometric series whose beginning index is other than zero. The value of \(\sum\limits_{n=r}^{\infty}z^n\) equals \(\frac{z^{r}}{ 1-z}\text{.}\) If we think of \(z\) as the “ratio” by which any term of the series is multiplied to generate successive terms, we note that the sum of a geometric series equals \(\frac{\;\text{ first term\; } }{\;1-\text{ ratio } \;}\text{,}\) provided \(|\text{ ratio } |\lt 1\text{.}\)
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The geometric series is used in the proof of Theorem 4.3.1, which is known as the ratio test. It is one of the most commonly used tests for determining the convergence or divergence of series. The proof is similar to the one used for real series, and we leave it for you to do.
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Example 4.3.8.

Show that \(\sum\limits_{n=0}^{\infty}\frac{(1-i)^n}{n!}\) converges.
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Solution.
Using the ratio test, we find that
\begin{align*} \lim_{n \to \infty}\frac{|{(1-i)^{n+1}}/{(n+1) !}|}{|{(1-i)^n}/{n!}|} \amp = \lim_{n \to \infty}\frac{n!|1-i|}{(n+1) !}=\lim_{n \to \infty}\frac{|1-i|}{n+1}\\ \amp = \lim_{n \to \infty}\frac{\sqrt{2}}{n+1}=0=L\text{.} \end{align*}
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Because \(L\lt 1\text{,}\) the series converges.
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Example 4.3.9.

Show that the series \(\sum\limits_{n=0}^{\infty}\frac{(z-i)^n}{2^n}\) converges for all values of \(z\) in the disk \(|z-i|\lt 2\) and diverges if \(|z-i|>2\text{.}\)
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Solution.
Using the ratio test, we see that
\begin{equation*} \lim_{n \to \infty}\frac{|{(z-i)^{n+1}}/{2^{n+1}}|}{|{(z-i)^n}/{2^n}|}= \lim_{n \to \infty}\frac{|z-i|}{2}=\frac{|z-i|}{2}=L\text{.} \end{equation*}
If \(|z-i|\lt 2\text{,}\) then \(L\lt 1\text{,}\) and the series converges. If \(|z-i|>2\text{,}\) then \(L>1\text{,}\) and the series diverges.
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Our next result, known as the root test, is slightly more powerful than the ratio test. Before we present this test, we need to discuss a rather sophisticated idea used with it—the limit supremum.
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Definition 4.3.10. Limit Supremum.

Let \(\{t_n\}\) be a sequence of positive real numbers. The limit supremum of the sequence (denoted \(\lim\limits_{n \to \infty} \sup \, t_n)\) is the smallest real number \(L\) having the property that for any \(\varepsilon>0\text{,}\) there are at most finitely many terms in the sequence that are larger than \(L+\varepsilon\text{.}\) If there is no such number \(L\text{,}\) then \(\lim\limits_{n \to \infty} \sup \, t_n=\infty\text{.}\)
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Example 4.3.11.

\textrm{} The limit supremum of the sequence
\begin{equation*} \{t_n\} =\{ 4.1,\;5.1,\;4.01,\;5.01,\;4.001,\;5.001, \ldots\} \text{{is} } \lim_{n \to \infty}\sup \, t_n=5\text{,} \end{equation*}
because if we set \(L=5\text{,}\) then for any \(\varepsilon>0\text{,}\) there are only finitely many terms in the sequence larger than \(L+\varepsilon =5+\varepsilon\text{.}\) Additionally, if \(L\) is smaller than 5, then by setting \(\varepsilon=5-L\text{,}\) we can find infinitely many terms in the sequence larger than \(L+\varepsilon\) (because \(L+\varepsilon=5)\text{.}\)
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Example 4.3.12.

The limit supremum of the sequence
\begin{equation*} \{t_n\} =\{ 1,\;2,\;3,\;1,\;2,\;3,\;1,\;2,\;3,\;1,\;2,\;3, \ldots \} \text{ is } \lim_{n \to \infty}\sup \, t_n=3\text{,} \end{equation*}
because if we set \(L=3\text{,}\) then for any \(\varepsilon>0\text{,}\) there are only finitely many terms (actually, there are none) in the sequence larger than \(L+\varepsilon=3+\varepsilon\text{.}\) Additionally, if \(L\) is smaller than 3, then by setting \(\varepsilon=\frac{3-L}{2}\) we can find infinitely many terms in the sequence larger than \(L+\varepsilon\) (because \(L+\varepsilon\lt 3)\text{,}\) as the following calculation shows:
\begin{equation*} L+\varepsilon=L+\frac{3-L}{2}=\frac{3+L}{2}=\frac{3}{2}+\frac{L}{2}\lt \frac{3 }{2}+\frac{3}{2}=3\text{.} \end{equation*}
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Example 4.3.13.

The limit supremum of the Fibonacci sequence
\begin{equation*} \{t_n\} =\{1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\ldots \} \text{ is } \lim_{n \to \infty}\sup \, t_n=\infty\text{.} \end{equation*}
(The Fibonacci sequence satisfies the relation \(t_n=t_{n-1}+t_{n-2}\) for \(n>2\text{.}\))
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The limit supremum is a powerful idea because the limit supremum of a sequence always exists, which is not true for the ordinary limit. However, Example 4.3.14 illustrates the fact that, if the limit of a sequence does exist, then it will be the same as the limit supremum.
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Example 4.3.14.

The sequence
\begin{align*} \{t_n\} \amp = \{1+\frac{1}{n}\}\\ \amp = \{2,\;1.5,\;1.3\overline{3},\;1.25,\;1.2,\;\ldots \} \text{ has } \lim_{n \to \infty}\sup \, t_n=1\text{.} \end{align*}
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We leave verification of this as an exercise.
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Proof.

Suppose first that \(L\lt 1\text{.}\) We can select a number \(r\) such that \(L\lt r\lt 1\text{.}\) By definition of the limit supremum, only finitely many terms in the sequence \(\{|\zeta _n|^{\frac{1}{n}}\}\) exceed \(r\) , so there exists a positive integer \(N\) such that for all \(n>N\) we have \(|\zeta_n|^{\frac{1}{n}}\lt r\text{.}\) That is, \(| \zeta _n|\lt r^n\) for all \(n>N\text{.}\) For \(r\lt 1\text{,}\) Theorem 4.3.1 implies that \(\sum\limits_{n=N+1}^{\infty}r^n\) converges. But then Theorem 4.1.24 implies that \(\sum \limits_{n=N+1}^{\infty}|\zeta _n|\) converges, and hence so does \(\sum\limits_{n=0}^{\infty}|\zeta _n|\) . Corollary 4.1.25 then guarantees that \(\sum\limits_{n=0}^{\infty}\zeta _n\) converges.
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Now suppose that \(L>1\text{.}\) We can select a number \(r\) such that \(1\lt r\lt L\text{.}\) Again, by definition of the limit supremum we conclude that \(|\zeta _n|^{\frac{1}{n}}>r\) for infinitely many \(n\text{.}\) But this condition means that \(|\zeta _n|>r^n\) for infinitely many \(n\text{,}\) and as \(r>1\text{,}\) this implies that \(\zeta _n\) does not converge to 0. By Theorem 4.1.17, \(\sum\limits_{n=0}^{\infty}\zeta _n\) does not converge.
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Note that, in applying either Theorem 4.3.7 or 4.3.15, if \(L=1\) the convergence or divergence of the series is unknown, and further analysis is required to determine the true state of affairs.
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Exercises Exercises

1.

Evaluate
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(a)
\(\sum\limits_{n=0}^{\infty}\frac{(1+i)^n}{2^n}\text{.}\)
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Solution.
By Theorem 4.3.1, \(\sum\limits_{n=0}^{\infty}\frac{(1+i)^n}{2^n}=\frac{1}{1-(\frac{1+i}{2})}=1+i\) (show the details), since \(|\frac{1+i}{2}|\lt 1\) (show this also).
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(b)
\(\sum\limits_{n=0}^{\infty}(\frac{1}{2+i})^n\text{.}\)
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2.

Show that \(\sum\limits_{n=0}^{\infty} \frac{(z+i)^n}{2^n}\) converges for all values of \(z\) in the disk \(D_2(-i) = \{z:|z+i|\lt 2\}\) and diverges if \(|z+i| > 2\text{.}\)
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3.

Is the series \(\sum\limits_{n=0}^{\infty}\frac{(4i)^n}{n!}\) convergent? Why or why not?
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Solution.
The series converges by the ratio test. Show the details.
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4.

Use the ratio test to show that the following series converge.
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(a)
\(\sum\limits_{n=0}^{\infty}(\frac{1+i}{2})^n\text{.}\)
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(b)
\(\sum\limits_{n=1}^{\infty}\frac{(1+i)^n}{n2^n}\text{.}\)
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(c)
\(\sum\limits_{n=1}^{\infty}\frac{(1+i)^n}{n!}\text{.}\)
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(d)
\(\sum\limits_{n=0}^{\infty}\frac{(1+i)^{2n}}{(2n+1)!}\text{.}\)
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5.

Use the ratio test to find a disk in which the following series converge.
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(a)
\(\sum\limits_{n=0}^{\infty}(1+i)^nz^n\text{.}\)
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Solution.
Converges in \(D_\frac{\sqrt{2}}{2}(0)\text{.}\)
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(b)
\(\sum\limits_{n=0}^{\infty}\frac{z^n}{(3+4i)^n}\text{.}\)
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(c)
\(\sum\limits_{n=0}^{\infty}\frac{(z-i)^n}{(3+4i)^n}\text{.}\)
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Solution.
Converges in \(D_{5}(i)\text{.}\)
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(d)
\(\sum\limits_{n=0}^{\infty}\frac{(z-3-4i)^n}{2^n}\text{.}\)
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6.

Establish the claim in the proof of Theorem 4.3.1 that, if \(|z|\lt 1\text{,}\) then \(\lim\limits_{n \to \infty}z^n=0\text{.}\)
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7.

In the geometric series, show that if \(|z|>1\text{,}\) then \(\lim\limits_{n \to \infty}|S_n|=\infty\text{.}\)
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Solution.
\(|S_n| = |\frac{1}{1-z} - \frac{z^n}{1-z}| \ge |\frac{z^n}{1-z}| - |\frac{1}{1-z}| = |z^n| \, |\frac{1}{1-z}| - |\frac{1}{1-z}|\text{.}\) Now use the fact that \(|z|>1\) to get the desired conclusion.
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Solution.
Mimic the argument most calculus texts give for real series, but replace \(|x|\) with \(|z|\text{.}\)
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11.

For \(|z|\lt 1\text{,}\) let \(f(z)=\sum\limits_{n=0}^{\infty}z^{(2^n)}=z+z^2+z^4+\cdots +z^{2^n}+\cdots\text{.}\) Show that \(f(z) =z+f(z^2)\text{.}\)
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Solution.
If \(f(z)=\sum\limits_{n=0}^{\infty }z^{(2^n)}\text{,}\) then \(f(z^2) = \sum\limits_{n=0}^{\infty}(z^2)^{(2^n)} = \sum\limits_{n=0}^{\infty}z^{(2\cdot 2^n)} = \sum\limits_{n=0}^{\infty}z^{(2^{n+1})} = \sum\limits_{n=1}^{\infty}z^{(2^n)}\text{.}\) Explain why the conclusion now follows, especially the second equality for \(f(z^2)\text{.}\)
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12.

This exercise makes interesting use of the geometric series.
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(a)
Use the formula for geometric series with \(z=re^{i\theta}\text{,}\) where \(r\lt 1\text{,}\) to show that
\begin{equation*} \sum_{n=0}^{\infty}z^n=\sum\limits_{n=0}^{\infty}r^ne^{in\theta} =\frac{1-r\cos \theta +ir\sin \theta}{1+r^2-2r\cos \theta} \end{equation*}
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(b)
Use part (a) to obtain
\begin{align*} \sum\limits_{n=0}^{\infty}r^n\cos n\theta \amp = \frac{1-r\cos \theta}{1+r^2-2r\cos \theta}, \text{ and }\\ \sum\limits_{n=0}^{\infty}r^n\sin n\theta \amp = \frac{r\sin \theta}{1+r^2-2r\cos \theta}\text{.} \end{align*}
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