absolutely convergent infinite product converges

Theorem.  An absolutely convergent (http://planetmath.org/AbsoluteConvergenceOfInfiniteProduct) infinite product

${\displaystyle \prod _{\nu =1}^{\mathrm{\infty }}}(1+c_{\nu })=(1+c_1)(1+c_2)(1+c_3)\mathrm{\cdots }$(1)

of complex numbers is convergent.

Proof.  We thus assume the convergence of the product (http://planetmath.org/Product)

${\displaystyle \prod _{\nu =1}^{\mathrm{\infty }}}(1+|c_{\nu }|)=(1+|c_1|)(1+|c_2|)(1+|c_3|)\mathrm{\cdots }$(2)

Let $\epsilon$ be an arbitrary positive number.  By the general convergence condition of infinite product, we have

when  $n\geqq$ certain $n_{\epsilon }$.  Then we see that

$|(1+c_{n+1})(1+c_{n+2})\mathrm{\cdots }(1+c_{n+p})-1|$$=|1+{\displaystyle \sum _{\nu =n+1}^{n+p}}{c}_{\nu }+{\displaystyle \sum _{\mu ,\nu }}{c}_{\mu }{c}_{\nu }+\mathrm{\dots }+c_{n+1}{c}_{n+2}\mathrm{\cdots }{c}_{n+p}-1|$

$\leqq 1+{\displaystyle \sum _{\nu =n+1}^{n+p}}|c_{\nu }|+{\displaystyle \sum _{\mu ,\nu }}|c_{\mu }||c_{\nu }|+\mathrm{\dots }+|c_{n+1}||c_{n+2}|\mathrm{\cdots }|c_{n+p}|-1$

as soon as  $n\geqq n_{\epsilon }$.  I.e., the infinite product (1) converges, by the same convergence condition.