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\newpage
\section{Power Series [4-5]}

We want to look at
\[ \sum_{n = 0}^\infty a_n z^n \tag{$*$} \]
$z \in \CC$, $a_n \in \CC$. (The case $\sum_{n =
0}^\infty a_n(z - z_0)^n$, $z_0$ fixed, can be
reduced to ($*$) by translation).

\begin{lemma}
    If $\sum_{n = 0}^\infty a_n z_1^n$ converges
    and $|z| < |z_1|$, then $\sum_{n = 0}^\infty
    a_n z^n$ \emph{converges absolutely}.
\end{lemma}

\begin{proof}
    Since $\sum_{n = 0}^\infty a_n z_1^n$
    converges, $a_n z_1^n \to 0$. Thus $\exists
    \,\, K > 0$ such that $|a_n z_1^n| \le K
    \,\,\forall\,\, n$. Then
    \begin{align*}
        |a_n z^n|
        &= |a_n z^n| \frac{|z_1^n|}{|z_1^n|} \\
        &\le K \ub{\left| \frac{z}{z_1}
        \right|^n}_{< 1}
    \end{align*}
    Since the geometric series
    \[ \sum_{n = 0}^\infty \left| \frac{z}{z_1}
    \right|^n \]
    converges, the lemma follows by comparison.
\end{proof}

\myskip
Using this lemma, we'll prove that every power
series has a \emph{radius of convergence}.

\begin{theorem}
    A power series either
    \begin{enumerate}[(1)]
        \item Converges absolutely for all $z$, or
        \item Converges absolutely for all $z$
            inside a circle $|z| = R$ and diverges
            for all $z$ outside it, or
        \item Converges for $z = 0$ only.
    \end{enumerate}
\end{theorem}

\begin{definition*}
    The circle $|z| = R$ is called the circle of
    convergence and $R$ is the radius of
    convergence. In (1) we agree that $R = \infty$
    and in (3) $R = 0$ (so $R \in [0, \infty]$).
\end{definition*}

\begin{proof}
    Let
    \[ S = \{x \in \RR : x \ge 0 \text{ and
    $\sum a_n x^n$ converges}\} \]
    Clearly $0 \in S$. By Lemma 4.1 if $x_1 \in
    S$, then $[0, x_1] \subset S$. If $S = [0,
    \infty)$ we have case (1). If not, there
    exists a finite supremum for $S$. Let $R =
    \sup S < \infty$, $R \ge 0$. If $R > 0$, we'll
    prove that if $|z_1| < R$, then $\sum a_n
    z_1^n$ converges absolutely. Pick $R_0$ such
    that
    \[ |z_1| < R_0 < R \]
    Then $R_0 \in S$ and the series converges for
    $z = R_0$. By Lemma 4.1, $\sum |a_n z_1^n|$
    converges. Finally we show that if $|z_2| >
    R$, then the series does not converge for
    $z_2$. Pick $R < R_0 < |z_2|$. If $\sum a_n
    z_2^n$ converges then by Lemma 4.1 $\sum a_n
    R_0^n$ would be convergent, which contradicts
    that $R = \sup S$.
\end{proof}

\myskip
The following lemma is useful for computing $R$:

\begin{lemma}
    If $\left| \frac{a_{n + 1}}{a_n} \right| \to
    l$ as $n \to \infty$, then $R = \frac{1}{l}$.
\end{lemma}

\begin{proof}
    By the ratio test we have absolute convergence
    if
    \[ \lim \left| \frac{a_{n + 1}}{a_n}
    \frac{z^{n + 1}}{z^n} \right| < 1 \]
    so if $|z| < \frac{1}{l}$ we have absolute
    convergence. If $|z| > \frac{1}{l}$, the
    series diverges, again by ratio test.
\end{proof}

\begin{remark*}
    One can also use the root test to get that if
    $|a_n|^{1/n} \to l$, then $R = \frac{1}{l}$.
\end{remark*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $\sum_{n = 0}^\infty \frac{z^n}{n!}$.
        \[ \left| \frac{a_{n + 1}}{a_n} \right| =
        \frac{n!}{(n + 1)!} = \frac{1}{n + 1} \to
        0 = l \implies R = \infty \]
    \item Geometric series, $\sum_{n = 0}^\infty
        z^n$. $R = 1$. Note that at $|z| = 1$ we
        have divergence.
    \item $\sum_{n = 0}^\infty n! z^n$.
        \[ \left| \frac{a_{n + 1}}{a_n} \right| =
        \frac{(n + 1)!}{n!} = n + 1 \to \infty
        \implies R = 0 \]
    \item $\sum_{n = 1}^\infty \frac{z^n}{n}$, $R
        = 1$. (for $z = 1$ it diverges (harmonic
        series)) What happens for $|z| = 1$ and $z
        \neq 1$? Consider $\sum_{n = 1}^\infty
        \frac{z^n}{n} (1 - z)$. Then
        \begin{align*}
            s_N = \sum_{n = 1}^N \left( \frac{z^n
            - z^{n + 1}}{n} \right) \\
            &= \sum_{n = 1}^N \frac{z^n}{n} -
            \sum_{n = 1}^N \frac{z^{n + 1}}{n} \\
            &= \sum_{n = 1}^N \frac{z^n}{n} -
            \sum_{n = 2}^{N + 1} \frac{z^n}{n - 1}
            \\
            &= z - \frac{z^{N + 1}}{N} + \sum_{n =
            2}^N z^n \left( -\frac{1}{n(n - 1)} \right)
        \end{align*}
        if $|z| = 1$, then $\frac{z^{N + 1}}{N}
        \to 0$ as $N \to \infty$ and $\sum
        \frac{1}{n(n - 1)}$ converges, so $s_N$
        \emph{converges}.
    \item $\sum_{n = 1}^\infty \frac{z^n}{n^2}$,
        $R = 1$ but converges for all $z$ with
        $|z| = 1$.
\end{enumerate}

\subsubsection*{Conclusion}

In principle nothing can be said about $|z| = R$
and each case has to be discussed separately.
Within the radius of convergence ``life is
great''. Power series behave as if ``they were
polynomials''.
