convergence condition of infinite productConvergenceConditionOfInfiniteProduct2004-09-21 17:55:272008-12-28 16:27:02Theoremconvergence/divergence for an infinite productinfinite productvalue of infinite productCauchy sequence% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}Let us think the sequence \,$u_1,\,u_1u_2,\,u_1u_2u_3,\,\ldots$\, In the complex analysis, one often uses the definition of the convergence of an {\em infinite product}\, $\displaystyle\prod_{k = 1}^{\infty}u_k$\, where the case\, $\displaystyle\lim_{k\to\infty}u_1u_2 \ldots u_k = 0$\, is excluded.\, Then one has the
\begin{thmplain}\, The infinite product $\displaystyle\prod_{k = 1}^{\infty}u_k$ of the non-zero complex numbers\, $u_1$, $u_2$, ... is convergent iff for every positive number $\varepsilon$ there exists a positive number $n_\varepsilon$ such that the condition
$$\vert u_{n+1}u_{n+2} \ldots u_{n+p}-1 \vert < \varepsilon \quad \forall \,p\in\mathbb{Z}_+$$
is true as soon as\, $n \geqq n_\varepsilon$.
\end{thmplain}
\textbf{Corollary.}\, If the infinite product converges, then we necessarily have\, $\displaystyle\lim_{k\to\infty}u_k = 1$. (Cf. the necessary condition of convergence of series.)
When the infinite product converges, we say that the {\em value of the infinite product} is equal to $\displaystyle\lim_{k\to\infty} u_1u_2 \ldots u_k$.