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\subsubsection*{Absolute Convergence}

\begin{definition*}
    Take $a_n \in \CC$. If $\sum_{n = 1}^\infty
    |a_n|$ is convergent, then the series is
    called \emph{absolutely convergent}.
\end{definition*}

\begin{note*}
    Since $|a_N| \ge 0$, we can use the previous
    tests to check absolute convergence. This is
    particularly useful for $a_n \in \CC$.
\end{note*}

\begin{theorem}
    If $\sum a_n$ is absolutely convergent, then
    it is convergent.
\end{theorem}

\begin{proof}
    Suppose first $a_n \in \RR$. Let
    \[ v_N = \begin{cases}
        a_n & \text{if $a_n \ge 0$} \\
        0 & \text{if $a_n < 0$}
    \end{cases} \]
    \[ w_n = \begin{cases}
        0 & \text{if $a_n \ge 0$} \\
        -a_n & \text{if $a_n < 0$}
    \end{cases} \]
    \[ v_n = \frac{|a_n| + a_n}{2}, \qquad w_n =
    \frac{|a_n| - a_n}{2} \]
    Clearly, $v_n, w_n \ge 0$. Note $a_n = v_n -
    w_n$, and $|a_n| = v_n + w_n \ge v_n, w_n$. So
    if $\sum |a_n|$ converges, by comparison $\sum
    v_n$, $\sum w_n$ also converge, hence $\sum
    a_n$ converges. If $a_n \in \CC$, then $a_n =
    x_n + iy_n$. Now $|x_n|, |y_n| \le |a_n|$, so
    $\sum x_n$ and $\sum y_n$ are absolutely
    convergent, hence $\sum x_n$ and $\sum y_n$
    converge. Since $a_n = x_n + iy_n$ we have
    that $\sum a_n$ converges as well.
\end{proof}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $\sum \frac{(-1)^{n + 1}}{n}$ converges
        but is \emph{not} absolutely convergent.
    \item $\sum_{n = 1}^\infty \frac{z^n}{2^n}$
        for $z \in \CC$, then if $|z| < 2$ we have
        absolute convergence. If $|z| \ge 2$,
        $\left| \frac{z}{2} \right|^n \ge 1$, so
        $a_n$ does not tend to 0, hence the series
        diverges.
\end{enumerate}

\begin{definition*}
    If $\sum a_n$ converges, but $\sum |a_n|$ does
    not, it is said sometimes, that $\sum a_n$ is
    \emph{conditionally} convergent.
\end{definition*}

\noindent
``conditional'': because the sum to which the
series converge is conditional on the order in
which elements of the sequence are taken. If
\emph{rearranged}, the sum is altered.

\begin{example*}
    (Example Sheet 1, Q7)
    \begin{enumerate}[(i)]
        \item $1 - \frac{1}{2} + \frac{1}{3} -
            \frac{1}{4} + \cdots$.
        \item $1 + \frac{1}{3} - \frac{1}{2} +
            \frac{1}{5} + \frac{1}{7} -
            \frac{1}{4} + \frac{1}{9} + \cdots$.
    \end{enumerate}
    Let $s_n$ be the partial sum of (i) and $t_n$
    the partial sum of (ii). Then $s_n \to s > 0$,
    and $t_n \to \frac{3s}{2}$.
\end{example*}

\noindent
Rearrangement:

\begin{definition*}
    Let $\sigma$ be a bijection of the positive
    integers,
    \[ a'_n = a_{\sigma(n)} \]
    is a rearrangement.
\end{definition*}

\begin{theorem}
    If $\sum a_n$ is absolutely convergent, every
    series consisting of the same terms in any
    order (i.e. a rearrangement) has the
    \emph{same sum}.
\end{theorem}

\begin{proof}
    We do the proof first for $a_n \in \RR$. Let
    $\sum a'_n$ be a rearrangement of $\sum a_n$.
    Let $s_n = \sum_{j = 1}^n a_j$ and $t_n =
    \sum_{j = 1}^n a'_j$, $s = \sum_{j = 1}^\infty
    a_j$. Suppose first that $a_n \ge 0$. Given
    $n$, we can find $q$ such that $sq$ satisfies
    \[ t_n \le sq \le s \]
    Now since $t_n$ is an increasing sequence
    bounded above, $t_n \to t$. Clearly $t \le s$.
    But by symmetry, $s \le t$, hence $t = s$. \\
    If $a_n$ has any sign, consider $v_n$ and
    $w_n$ from theorem 1.13. Consider $\sum a'_n$,
    $\sum v'_n$ and $\sum w'_n$. Since $\sum
    |a_n|$ converges, both $\sum v_n$ and $\sum
    w_n$ converge. Use that $v_n, w_n \ge 0$ to
    deduce that $\sum v'_n = \sum v_n$ and $\sum
    w'_n = \sum w_n$. But $a_n = v_n - w_n$ hence
    $\sum a_n = \sum a'_n$. \\
    For the case $a_n \in \CC$, write $a_n = x_n +
    iy_n$. Since $|x_n|, |y_n| \le |a_n|$, we have
    that $\sum x_n$ and $\sum y_n$ are absolutely
    convergent. By the previous case, $\sum x'_n =
    \sum x_n$ and $\sum y'_n = \sum y_n$ since
    $a'_n = x'_n + iy'_n \implies \sum a_n = \sum
    a'_n$.
\end{proof}