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\noindent {\bf Question} Unlike sequences, the convergence of
series whose terms are products and quotients of convergent series
does not necessarily follow. Exploring this phenomenon is the
purpose of this example. Construct examples of each of the
following, or prove that no such example exists:
\begin{enumerate}
\item convergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$  with positive terms for which the series of products
$\sum_{n=0}^\infty a_n\: b_n$ converges;
\item divergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$ with positive terms for which the series of products
$\sum_{n=0}^\infty a_n\: b_n$ diverges;
\item divergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$ with positive terms for which the series of products
$\sum_{n=0}^\infty a_n\: b_n$ converges;
\item convergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$ with positive terms for which the series of quotients
$\sum_{n=0}^\infty \frac{a_n}{b_n}$ diverges;
\item convergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$ with positive terms for which the series of quotients
$\sum_{n=0}^\infty \frac{a_n}{b_n}$ converges;
\item divergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$ with positive terms for which the series of quotients
$\sum_{n=0}^\infty \frac{a_n}{b_n}$ diverges;
\item divergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty
b_n$ with positive terms for which the series of quotients
$\sum_{n=0}^\infty \frac{a_n}{b_n}$ converges;
\end{enumerate}
\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item All we need are two convergent
series.  For instance, take $a_n = (0.5)^n$ and $b_n =(0.3)^n$ for
all $n\ge 0$.  Then, $\sum_{n=0}^\infty a_n$, $\sum_{n=0}^\infty
b_n$, and $\sum_{n=0}^\infty a_n\: b_n$ are all convergent
geometric series.
\item take $a_n =1$ for all $n\ge 0$ and $b_n = 1$ for all $n\ge 0$.
Then, both $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ are
both divergent geometric series, as is $\sum_{n=0}^\infty a_n\:
b_n$ (since $a_n\: b_n =1$ for all $n\ge 0$).
\item for this one, let's take $a_n =b_n =\frac{1}{n}$ for all $n\ge
1$.  Then, both $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty
b_n$ are the harmonic series, and hence divergent.  However, the
series of products $\sum_{n=1}^\infty a_n\: b_n =\sum_{n=1}^\infty
\frac{1}{n^2}$ is convergent, by the note below.
\item take any convergent series, for example $\sum_{n=0}^\infty
(0.5)^n$, and set $a_n =b_n =(0,5)^n$.  Then, the series of
quotients is $\sum_{n=0}^\infty \frac{a_n}{b_n} =\sum_{n=0}^\infty
1$, which diverges.
\item here, we can take $a_n =\frac{1}{n^2}$ and $b_n =\frac{1}{n^4}$
for $n\ge 1$.  Then, both $\sum_{n=1}^\infty a_n
=\sum_{n=1}^\infty \frac{1}{n^2}$ and $\sum_{n=1}^\infty b_n
=\sum_{n=1}^\infty \frac{1}{n^4}$ converge by the note below, as
does the series of quotients, as $\frac{a_n}{b_n} =\frac{1}{n^2}$.
\item let's use geometric series again: both of $\sum_{n=0}^\infty a_n
=\sum_{n=0}^\infty 6^n$ and $\sum_{n=0}^\infty b_n
=\sum_{n=0}^\infty 2^n$ are divergent geometric series, and the
series of quotients $\sum_{n=0}^\infty \frac{a_n}{b_n}
=\sum_{n=0}^\infty 3^n$ is also a divergent geometric series.
\item $\sum_{n=1}^\infty a_n =\sum_{n=1}^\infty 1$ and
$\sum_{n=1}^\infty b_n =\sum_{n=1}^\infty n^2$ both diverge, but
the corresponding sequence of quotients $\sum_{n=1}^\infty
\frac{a_n}{b_n} =\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
\end{enumerate}

\noindent Note

\noindent The series $\sum_{n=1}^\infty \frac{1}{n^s}$ converges
if and only if $s >1$.

\medskip
\noindent For $s =1$, this series is called the {\bf harmonic
series}, and we can prove directly that it diverges.  Note that
$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$, that $\frac{1}{5} +
\cdots + \frac{1}{8} > 4\frac{1}{8} = \frac{1}{2}$, and in general
that
\[ \frac{1}{2^{k-1} +1} + \frac{1}{2^{k-1} +2} +\cdots +\frac{1}{2^k}
> 2^{k-1}\frac{1}{2^k} =\frac{1}{2}. \]
Hence, the $(2^k)^{th}$ partial sum $S_{2^k}$ satisfies $S_{2^k}
>1 +k\frac{1}{2}$.  Since the terms in the harmonic series are all
positive, the sequence of partial sums is monotonically
increasing, and by the calculation done the sequence of partial
sums is unbounded, and so the sequence of partial sums diverges.
Hence, the harmonic series diverges.

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