% [** F1: Using \lim for limit on this page]
to be \emph{not absolutely convergent} or \emph{conditionally convergent}. To this
latter class belong some convergent alternating series. For example,
the series
\[
1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \frac{1}{5^5} - \dotsb
\]
is \emph{absolutely convergent}, since the series~$(C)$, p.~217, % [** F1: Hard-coded page number]
namely,
\[
1 + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \frac{1}{5^5} + \dotsb
\]
is convergent. The series
\[
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dotsb
\]
is \emph{conditionally} convergent, since the harmonic series
\[
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dotsb
\]
is divergent.

\textit{A series with terms of different signs is convergent if the series deduced
from it by making all the signs positive is convergent}.

The proof of this theorem is omitted.

% [** F1: Hard-coded page number below]
Assuming that the ratio test on p.~219 holds without placing any
restriction on the signs of the terms of a series, we may summarize
our results in the following %[**P3: : ?]

\paragraph{General directions for testing the series}
\[
u_1 + u_2 + u_3 + u_4 + \dotsb + u_n + u_{n+1} + \dotsb .
\]

\textit{When it is an alternating series whose terms never increase in numerical
value, and if
\[
\lim_{n=\infty} u_n = 0,
\]
then the series is convergent}.

\textit{In any series in which the above conditions are not satisfied, we determine
the form of $u_n$ and $u_{n+1}$ and calculate the limit}
\[
\lim_{n=\infty} \left(\frac{u_{n+1}}{u_n}\right) = \rho.
\]
\begin{itemize}
\item[  I.] \textit{When $|\rho| < 1$, the series is absolutely convergent}.

\item[ II.] \textit{When $|\rho| > 1$, the series is divergent}.

\item[III.] \textit{When $|\rho| = 1$, there is no test, and we should compare the series
with some series which we know to be convergent, as}
\begin{align*}
% Use of \tag is a dirty trick....
&a + ar + ar^2 + ar^3 + \dotsb;\ r < 1, \tag{\text{\textit{geometric series}}} \\
&1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \dotsb;\ p > 1, \tag{\text{$p$~series}}
\end{align*}
\end{itemize}
