# absolutely convergent infinite product converges

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

 $\displaystyle\prod_{\nu=1}^{\infty}(1\!+\!c_{\nu})\;=\;(1\!+\!c_{1})(1\!+\!c_{% 2})(1\!+\!c_{3})\cdots$ (1)

of complex numbers is convergent.

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

 $\displaystyle\prod_{\nu=1}^{\infty}(1\!+\!|c_{\nu}|)\;=\;(1\!+\!|c_{1}|)(1\!+% \!|c_{2}|)(1\!+\!|c_{3}|)\cdots$ (2)

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

 $\displaystyle|(1\!+\!c_{n+1})(1\!+\!c_{n+2})\cdots(1\!+\!c_{n+p})-1|$ $\displaystyle=|1+\sum_{\nu=n+1}^{n+p}c_{\nu}+\sum_{\mu,\,\nu}c_{\mu}c_{\nu}+% \ldots+c_{n+1}c_{n+2}\cdots c_{n+p}-1|$ $\displaystyle\leqq 1+\sum_{\nu=n+1}^{n+p}|c_{\nu}|+\sum_{\mu,\,\nu}|c_{\mu}||c% _{\nu}|+\ldots+|c_{n+1}||c_{n+2}|\cdots|c_{n+p}|-1$ $\displaystyle=|(1\!+\!|c_{n+1}|)(1\!+\!|c_{n+2}|)\cdots(1\!+\!|c_{n+p}|)-1|<% \varepsilon\qquad\forall\;p\in\mathbb{Z}_{+}$

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

Title absolutely convergent infinite product converges AbsolutelyConvergentInfiniteProductConverges 2013-03-22 18:41:15 2013-03-22 18:41:15 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 40A05 msc 30E20 convergence of absolutely convergent infinite product AbsoluteConvergenceImpliesConvergenceForAnInfiniteProduct AbsoluteConvergenceOfInfiniteProductAndSeries