Definition 7.4.1.
A power series centered at \(z_0\) is a series of the form
\begin{equation*}
\s c_k \left( z - z_0 \right)^k
\end{equation*}
where \(c_0, c_1, c_2, \ldots \in \C\text{.}\)
Section 7.4 Regions of Convergence
For the remainder of this chapter (indeed, this book) we concentrate on some very special series of functions.
Example 7.4.2.
A slight modification of Example 7.3.8 gives a fundamental power series, namely the geometric series
\begin{equation*}
\sum_{ k \ge 0 } z^k \ = \ \frac 1 {1-z} \,\text{.}
\end{equation*}
So here \(z_0 = 0\) and \(c_k = 1\) for all \(k \ge 0\text{.}\) We note that, as in Example 7.3.8, this power series converges absolutely for \(|z| \lt 1\) and uniformly for \(|z| \le r\text{,}\) for any fixed \(r\lt 1\text{.}\) Finally, as in Example 7.2.9, the geometric series \(\sum_{ k \ge 0 } z^k\) diverges for \(|z| \ge 1\text{.}\)
A general power series has a very similar convergence behavior which, in fact, comes from comparing it to a geometric series.
Theorem 7.4.3.
Given a power series \(\s c_k ( z - z_0)^k\text{,}\) there exists a real number \(R \ge 0\) or \(R = \infty\text{,}\) such that
-
\(\s c_k ( z - z_0)^k\) converges absolutely and uniformly for \(|z-z_0| \le r\text{,}\) for any \(r \lt R\text{;}\)
We remark that this theorem says nothing about the convergence/divergence of \(\s c_k ( z - z_0)^k\) for \(|z-z_0| = R\text{.}\)
Definition 7.4.4.
The number \(R\) in Theorem 7.4.3 is called the radius of convergence of \(\s c_k ( z - z_0)^k\text{.}\) The open disk \(D[z_0,R]\) in which the power series converges absolutely is the region of convergence. (If \(R=\infty\) then this is \(\C\text{.}\))
In preparation for the proof of Theorem 7.4.3, we start with the following observation.
Proposition 7.4.5.
If \(\s c_k ( w - z_0)^k\) converges, then \(\s c_k ( z - z_0)^k\) converges absolutely whenever \(|z
- z_0| \lt |w - z_0|\text{.}\)
Proof.
Let \(r := |w - z_0|\text{.}\) If \(\s c_k ( w - z_0)^k\) converges then \(\lim_{ k \to \infty } c_k ( w - z_0)^k = 0\) and so this sequence of terms is bounded (by Exercise 7.5.6), say
\begin{equation*}
\left| c_k ( w - z_0)^k \right| \ = \ |c_k| \, r^k \ \le \ M \,\text{.}
\end{equation*}
Now if \(|z - z_0| \lt |w - z_0|\text{,}\) then
\begin{equation*}
\s \left| c_k ( z - z_0)^k \right| \ = \ \s |c_k| \, r^k \left(
\frac{ |z-z_0| }{ r } \right)^k \ \le \ M \s \left( \frac{
|z-z_0| }{ r } \right)^k \text{.}
\end{equation*}
The sum on the right-hand side is a convergent geometric sequence, since \(|z-z_0| \lt r\text{,}\) and so \(\s c_k ( z - z_0)^k\) converges absolutely by Corollary 7.2.6.
With this preparation, we can now prove Theorem 7.4.3.
Proof.
Consider the set
\begin{equation*}
S \ := \ \left\{ x \in \R_{ \ge 0 } : \, \s c_k \, x^k \text{
converges } \right\} \text{.}
\end{equation*}
(This set is nonempty since \(0 \in S\text{.}\))
If \(S\) is unbounded then \(\s c_k ( z - z_0)^k\) converges absolutely and uniformly for \(|z-z_0| \le r\text{,}\) for any \(r\) (and so this gives the \(R = \infty\) case of Theorem 7.4.3): choose \(x \in S\) with \(x > r\text{,}\) then Proposition 7.4.5 says that \(\s c_k \, r^k\) converges absolutely. Since \(\left| c_k ( z - z_0)^k \right| \le |c_k| r^k\text{,}\) we can now use Proposition 7.3.7.
If \(S\) is bounded, let \(R\) be its least upper bound. If \(R = 0\) then \(\s c_k ( z - z_0)^k\) converges only for \(z = z_0\text{,}\) which establishes Theorem 7.4.3 in this case.
Now assume \(R > 0\text{.}\) If \(|z-z_0| \lt R\) then (because \(R\) is a least upper bound for \(S\)) there exists \(r \in S\) such that
\begin{equation*}
|z-z_0| \lt r \le R \,\text{.}
\end{equation*}
Thus \(\s c_k ( w - z_0)^k\) converges for \(w=z_0+r\text{,}\) and so \(\s c_k ( z - z_0)^k\) converges absolutely by Proposition 7.4.5. This finishes (a).
If \(|z-z_0| \le r\) for some \(r \lt R\text{,}\) again we can find \(x \in S\) such that \(r \lt x \le R\text{.}\) Then \(\s |c_k| \, r^k\) converges by Proposition 7.4.5, and so \(\s c_k ( z - z_0)^k\) converges absolutely and uniformly for \(|z-z_0|\le r\) by Proposition 7.3.7. This proves (b).
Finally, if \(|z-z_0| > R\) then there exists \(r \notin
S\) such that
\begin{equation*}
R \le r \lt |z-z_0| \,\text{.}
\end{equation*}
But \(\s c_k \, r^k\) diverges, so (by the contrapositive of Theorem 7.2.16) \(\s |c_k| \, r^k\) diverges, and so (by the contrapositive of Proposition 7.4.5) \(\s c_k ( z - z_0)^k\) diverges, which finishes (c).
Corollary 7.4.6.
If \(\lim_{ k \to \infty } \sqrt[k]{ |c_k| }\) exists then the radius of convergence of \(\s c_k ( z - z_0)^k\) equals
\begin{equation*}
R = \begin{cases}\infty \amp \text{ if } \lim_{ k \to
\infty } \sqrt[k]{ |c_k| } = 0 \, , \\ \frac{ 1 }{ \lim_{ k
\to \infty } \sqrt[k]{ |c_k| } } \amp \text{ otherwise. }
\end{cases}
\end{equation*}
Proof.
Given \(R\) as in the statement of the corollary, it suffices (by Theorem 7.4.3) to show that \(\s c_k ( z - z_0)^k\) converges for \(|z-z_0| \lt R\) and diverges for \(|z-z_0| > R\text{.}\)
Suppose \(r := |z-z_0| \lt R\text{.}\) Since \(\lim_{ k \to \infty } \sqrt[k]{ |c_k| } = \frac 1
R\) and \(\frac{ 2 }{ R+r } > \frac 1 R\text{,}\) there exists \(N\) such that \(\sqrt[k]{ |c_k| } \lt \frac{
2 }{ R+r }\) for \(k \ge N\text{.}\) For those \(k\) we then have
\begin{equation*}
\left| c_k (z-z_0)^k \right| \ = \ |c_k| |z-z_0|^k \ = \ \left(
\sqrt[k]{ |c_k| } \, r \right)^k \ \lt \ \left( \frac{ 2r
}{ R+r } \right)^k
\end{equation*}
and so \(\sum_{ k=N }^\infty c_k (z-z_0)^k\) converges (absolutely) by Proposition 7.3.7, because \(\frac{ 2r }{ R+r } \lt 1\) and thus \(\s \left( \frac{ 2r }{ R+r } \right)^k\) converges as a geometric series. Thus \(\s c_k (z-z_0)^k\) converges.
Now suppose \(r = |z-z_0| > R\text{.}\) Again because \(\lim_{ k \to \infty } \sqrt[k]{ |c_k| } =
\frac 1 R\) and now \(\frac{ 2 }{ R+r } \lt \frac 1 R\text{,}\) there exists \(N\) such that \(\sqrt[k]{ |c_k| } > \frac{
2 }{ R+r }\) for \(k \ge N\text{.}\) For those \(k\text{,}\)
\begin{equation*}
\left| c_k (z-z_0)^k \right| \ = \ \left( \sqrt[k]{ |c_k| } \,
r \right)^k \ > \ \left( \frac{ 2r }{ R+r } \right)^k \ > \
1 \, \text{,}
\end{equation*}
and so the sequence \(c_k (z-z_0)^k\) cannot converge to 0. Subsequently (by Corollary 7.2.8), \(\s c_k (z-z_0)^k\) diverges.
You might remember this corollary from calculus, where it goes by the name root test. Its twin sister, the ratio test, is the subject of Exercise 7.5.32.
Example 7.4.7.
For the power series \(\ds \s k \, z^k\) we compute
\begin{equation*}
\lim_{ k \to \infty } \sqrt[k]{ |c_k| } \ = \ \lim_{ k \to
\infty } \sqrt[k]{ k } \ = \ \lim_{ k \to \infty } e^{ \frac 1
k \ln(k) } \ = \ e^{ \lim_{ k \to \infty } \frac { \ln(k) } k }
\ = \ e^0 \ = \ 1 \, \text{,}
\end{equation*}
and Corollary 7.4.6 gives the radius of convergence 1. (Alternatively, we can argue by differentiating the geometric series.)
Example 7.4.8.
Consider the power series \(\ds \sum_{ k \ge 0 } \frac{ 1 }{
k! } \, z^k \text{.}\) Since
\begin{equation*}
\lim_{ k \to \infty } \left| \frac{ c_{ k+1 } }{ c_{ k } }
\right| \ = \ \lim_{ k \to \infty } \frac{ k! }{ (k+1)! } \ = \
\lim_{ k \to \infty } \frac{ 1 }{ k+1 } \ = \ 0 \, \text{,}
\end{equation*}
the ratio test (Exercise 7.5.32) implies that the radius of convergence of \(\sum_{ k \ge 0 }
\frac{ 1 }{ k! } \, z^k\) is \(\infty\text{,}\) and so the power series converges absolutely in \(\C\text{.}\)
1
In the next chapter, we will see that this power series represents the exponential function.
Corollary 7.4.9.
Suppose the power series \(\s c_k ( z - z_0 )^k\) has radius of convergence \(R>0\text{.}\) Then the series represents a function that is continuous on \(D[z_0, R]\text{.}\)
Proof.
Given any point \(w \in D[z_0, R]\text{,}\) we can find \(r\lt R\) such that \(w \in D[z_0, r]\) (e.g., if \(R\ne\infty\) then \(r = \frac{ |w-z_0| + R }{ 2 }\) will do the trick). Theorem 7.4.3 says that \(\s c_k ( z - z_0 )^k\) converges uniformly in \(D[z_0, r]\text{,}\) and so Proposition 7.3.4 implies that the power series is continuous in \(D[z_0, r]\text{,}\) and so particularly at \(w\text{.}\)
Corollary 7.4.10.
Suppose the power series \(\s c_k ( z - z_0 )^k\) has radius of convergence \(R>0\) and \(\gamma\) is a piecewise smooth path in \(D[z_0, R]\text{.}\) Then
\begin{equation*}
\int_\gamma \ \s c_k ( z - z_0 )^k \, \diff{z} \ = \ \s c_k
\int_\gamma ( z - z_0 )^k \, \diff{z} \, \text{.}
\end{equation*}
In particular, if \(\gg\) is closed then \(\ds \int_\gg \ \s c_k ( z -
z_0 )^k \, \diff{z} \ = \ 0 \text{.}\)
Proof.
Let \(r := \max_{ z \in \gg } \left| \gamma(z) -z_0 \right|\) (whose existence is guaranteed by Theorem A.0.1). Then \(\gg \subset \overline D[z_0, r]\) and \(r\lt R\text{.}\) Theorem 7.4.3 says that \(\s c_k \left( z - z_0 \right)^k\) converges uniformly in \(\overline D[z_0, r]\text{,}\) and so Proposition 7.3.6 allows us to switch integral and summation.
The last statement follows now with Exercise 4.5.15.
These corollaries will become extremely useful in the next chapter.
