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manipulating convergent series

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\begin{document}
The \PMlinkescapetext{terms} of the series in the following theorems are supposed to be either real or complex numbers.

\begin{thmplain}
 \, If the series\, $a_1+a_2+\cdots$\, and\, $b_1+b_2+\cdots$\, converge and have the sums $a$ and $b$, respectively, then also the series
\begin{align}
           (a_1+b_1)+(a_2+b_2)+\cdots
\end{align}
converges and has the sum $a\!+\!b$.
\end{thmplain}

{\em Proof.} \,The $n^\mathrm{th}$ partial sum of (1) has the limit
$$\lim_{n\to\infty}\sum_{j = 1}^n(a_j+b_j) = 
\lim_{n\to\infty}\sum_{j = 1}^na_j+\lim_{n\to\infty}\sum_{j = 1}^nb_j = a\!+\!b.$$

\begin{thmplain}
 \, If the series\, $a_1+a_2+\cdots$\, converges having the sum $a$ and if $c$ is any \PMlinkescapetext{constant}, then also the series
\begin{align}
              ca_1+ca_2+\cdots
\end{align}
converges and has the sum $ca$.
\end{thmplain}

{\em Proof.} \,The $n^\mathrm{th}$ partial sum of (2) has the limit
$$\lim_{n\to\infty}\sum_{j = 1}^nca_j = c\lim_{n\to\infty}\sum_{j = 1}^na_j = 
ca.$$

\begin{thmplain}
 \, If the \PMlinkescapetext{terms} of any converging series
\begin{align}
a_1+a_2+a_3+\cdots
\end{align}
are grouped arbitrarily without changing their \PMlinkescapetext{order}, then the resulting series
\begin{align}
(a_1+\cdots+a_{m_1})+(a_{m_1+1}+\cdots+a_{m_2})+(a_{m_2+1}+\cdots+a_{m_3})+\cdots
\end{align}
also converges and its sum equals to the sum of (3).
\end{thmplain}

{\em Proof.} \,Since all the partial sums of (4) are simultaneously partial sums of (3), they have as limit the sum of the series (3).
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