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absolutely convergent infinite product converges

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\begin{document}
\textbf{Theorem.}\; An \PMlinkname{absolutely convergent}{AbsoluteConvergenceOfInfiniteProduct} infinite product
\begin{align}
\prod_{\nu=1}^\infty(1\!+\!c_\nu) \;=\; (1\!+\!c_1)(1\!+\!c_2)(1\!+\!c_3)\cdots
\end{align}
of complex numbers is convergent.\\

{\em Proof.}\, We thus assume the convergence of the \PMlinkname{product}{Product}
\begin{align}
\prod_{\nu=1}^\infty(1\!+\!|c_\nu|) \;=\; (1\!+\!|c_1|)(1\!+\!|c_2|)(1\!+\!|c_3|)\cdots
\end{align}
Let $\varepsilon$ be an arbitrary positive number.\, By the general convergence condition of infinite product, we have
$$|(1\!+\!|c_{n+1}|)(1\!+\!|c_{n+2}|)\cdots(1\!+\!|c_{n+p}|)-1| < \varepsilon 
\quad \forall\; p \in \mathbb{Z}_+$$
when\, $n \geqq$ certain $n_\varepsilon$.\, Then we see that
\begin{align*}
|(1\!+\!c_{n+1})(1\!+\!c_{n+2})\cdots(1\!+\!c_{n+p})-1| 
& =    |1+\sum_{\nu=n+1}^{n+p}c_\nu+\sum_{\mu,\,\nu}c_\mu c_\nu+\ldots+c_{n+1}c_{n+2}\cdots c_{n+p}-1|\\
& \leqq 1+\sum_{\nu=n+1}^{n+p}|c_\nu|+\sum_{\mu,\,\nu}|c_\mu||c_\nu|+\ldots+|c_{n+1}||c_{n+2}|\cdots|c_{n+p}|-1\\
& = |(1\!+\!|c_{n+1}|)(1\!+\!|c_{n+2}|)\cdots(1\!+\!|c_{n+p}|)-1| < \varepsilon 
\qquad \forall\; p \in \mathbb{Z}_+
\end{align*}
as soon as\, $n \geqq n_\varepsilon$.\; I.e., the infinite product (1) converges, by the same convergence condition.

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