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\noindent {\bf Question}
\begin{enumerate}
\item Show that, if $\sum_{n=0}^\infty a_n$ converges and if
$\sum_{n=0}^\infty b_n$ diverges, then the series of sums
$\sum_{n=0}^\infty (a_n +b_n)$ diverges.
\item Show that, if $\sum_{n=0}^\infty a_n$ diverges and if $c\ne 0$,
then the series of multiples $\sum_{n=0}^\infty c\: a_n$ diverges.
\end{enumerate}
\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item we argue by contradiction: suppose that $\sum_{n=0}^\infty (a_n
+b_n)$ converges.  Since $\sum_{n=0}^\infty a_n$ converges by
assumption, the arithmetic series, yields that their difference
also converges.  However, their difference is $\sum_{n=0}^\infty
(a_n +b_n -a_n) =\sum_{n=0}^\infty b_n$, which diverges by
assumption, yielding the desired contradiction.
\item again we argue by contradiction: suppose that the series of
multiples $\sum_{n=0}^\infty c a_n$ converges.  Then, the sequence
$\{ T_k =\sum_{n=0}^k c a_n\}$ of partial sums converges.  Note
though that $T_k =\sum_{n=0}^k c a_n =c \sum_{n=0}^k a_n =c S_k$,
where $S_k$ is the $k^{th}$ partial sum of the series
$\sum_{n=0}^\infty a_n$. Since $\{ T_k\}$ converges, the sequence
$\{ \frac{1}{c} T_k =S_k\}$ also converges, by the arithmetic of
sequences (since the constant sequence $\{ \frac{1}{c}\}$
converges), and so the original series converges, a contradiction.
\end{enumerate}

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