2.4 Power Series

See: 9. Power Series from Real Analysis

a_{0} + a_{1} z + a_{2} z^{2} + \dots + a_{n} z^{n} + \dots

Where $a_{n}$ and and the variable $z$ are complex.

We can also consider:

\sum_{n = 0}^{\infty} a_{n} (z - z_{0})^{n}

Most trivial is a geometric series:

1 + z + \dots + z^{n - 1} = \frac{1 - z^{n}}{1 - z}

since $z^{n} \to 0$ for $| z | < 1$ and diverges otherwise. We conclude that the the geometric series converges to $\frac{1}{1 - z}$ for $| z | < 1$

Abel's Theorem

For every power series there exists a number $R$ , $0 < R < \infty$ called the radius of convergence with the following properties

The series converges absolutely for every $z$ with $| z | < R$ and converges uniformly.
If $| z | > R$ the series is divergent
In $| z | < R$ the sum of the series is an analytic function where the derivative has the same radius of convergence

Hadamard's Formula

\frac{1}{R} = lim_{n \to \infty} sup \sqrt[n]{| a_{n} |}

Proofs

Let $\frac{1}{R} = lim_{n \to \infty} sup \sqrt[n]{| a_{n} |}$
If $| z | < R$ we can find $ρ$ so that $| z | < ρ < R$ and $\frac{1}{ρ} > \frac{1}{R}$
By definition of the limsup we can find $N$ so that when $n > N$ ,

\begin{aligned} \sqrt[n]{| a_{n} |} \leq \frac{1}{R} < \frac{1}{ρ} & ⟹ \sqrt[n]{| a_{n} |} | z | < \frac{| z |}{ρ} \\ ⟹ | a_{n} | | z |^{n} < \frac{| z |^{n}}{ρ^{n}} \end{aligned}

$\sum_{n = 0}^{\infty} \frac{| z |^{n}}{ρ^{n}}$ converges since it's a geometric series and $| \frac{| z |}{ρ} | < 1 ⟹ \sum_{}^{} | a_{n} z^{n} |$ is absolutely convergent.
and by M-test it converges uniformly
3.
Lemma: $lim_{n \to \infty} \sqrt[n]{n} = 1$
Binomial expansion:
$(1 + δ_{n})^{n} = 1 + n δ_{n} + \frac{n (n - 1)}{2} δ_{n}^{2} + \frac{n (n - 1) (n - 2)}{6} δ_{n}^{3} + \dots$

\begin{aligned} \sqrt[n]{n} = 1 + δ_{n} & ⟹ n = (1 + δ_{n})^{n} > 1 + \frac{1}{2} n (n - 1) δ_{n}^{2} \\ ⟹ n > 1 + \frac{1}{2} n (n - 1) δ_{n}^{2} \\ ⟹ δ_{n}^{2} < \frac{2}{n} ⟹ δ_{n} \to 0 \\ ⟹ \sqrt[n]{n} \to 1 \end{aligned}

Proof

\underset{n \to \infty}{lim sup} \sqrt[n]{| n \cdot a_{n} |} = \underset{n \to \infty}{lim sup} \sqrt[n]{| a_{n} |}

| z | < R ⟹ f (z) = \sum_{0}^{\infty} a_{n} z^{n} = \underset{S_{n} (z)}{\underset{⏟}{a_{0} + \dots + a_{n} z^{n}}} + \underset{R_{n} (z)}{\underset{⏟}{a_{n + 1} z^{n + 1} + \dots}}

f_{1} (z) = \sum_{1}^{\infty} n \cdot a_{n} z^{n - 1}

WTS $f^{'} (z) = f_{1} (z)$

\begin{array}{r} \frac{f (z) - f (z_{0})}{z - z_{0}} - f_{1} (z_{0}) = \frac{S_{n} (z) - S_{n} (z_{0})}{z - z} - s_{n}^{'} (z_{0}) - f_{1} (z_{0}) + \frac{R_{n} (z) - R_{n} (z_{0})}{z - z_{0}} \end{array}

take $| z_{0} | < ρ < R$

\begin{aligned} | \frac{R_{n} (z) - R_{n} (z_{0})}{z - z_{0}} | & = | \frac{\sum_{k = n + 1}^{\infty} a_{n} z^{k} - \sum_{k = n + 1}^{\infty} a_{k} z_{0}^{k}}{z - z_{0}} | \\ = | \frac{\sum_{k = n + 1}^{\infty} a_{k} (z^{k} - z_{0}^{k})}{b} | \\ = \sum_{k = n + 1}^{\infty} | a_{k} (z^{k - 1} + z^{k - 2} z_{0} + \dots + z_{0}^{k - 1}) | \\ \leq \sum_{k = n}^{\infty} k | a_{k} | ρ^{k - 1} \end{aligned}

Which converges $\underset{k \to \infty}{lim sup} \sqrt[k]{k | a_{k} |} = \frac{1}{R}$

For $ρ < R$ ,
$ρ \cdot \frac{1}{R} < 1$ .

By the root test, the series $\sum k | a_{k} | ρ^{k - 1}$ converges

$⟹ \exists N, n > N ⟹ | \frac{R_{n} (z) - R_{n} (z_{0})}{z - z_{0}} | < \frac{ε}{3}$

\begin{array}{r} \frac{f (z) - f (z_{0})}{z - z_{0}} = \frac{S_{n} (z) - S_{n} (z_{0}) + (R_{n} (z) - R_{n} (z_{0}))}{z - z_{0}} \end{array}

s_{n}^{'} (z) = \sum_{k = 1}^{n} k a_{k} (z)^{k - 1}, f_{1} (z) = \sum_{k = 1}^{\infty} k a_{k} (z)^{k - 1} .

For $| z | \leq ρ < R$ ,

| a_{k} (z)^{k - 1} | \leq k | a_{k} | ρ^{k - 1}

Since $\sum k | a_{k} | ρ^{k - 1}$ converges the M-test gives uniform convergence of $s_{n}^{'} (z)$ to $f_{1} (z)$ .

Hence, for any $ε > 0$ ,

\exists n_{1} \forall n \geq n_{1} : | s_{n}^{'} (z_{0}) - f_{1} (z_{0}) | < ε / 3.

and by the definition of the derivative

| \frac{s_{n} (z) - s_{n} (z_{0})}{z - z_{0}} - s_{n}^{'} (z_{0}) | < \frac{ε}{3}

Combining these we get what we want.