# Proof of uniform convergence of an infinite product

I am reading the book Complex Analysis: An Invitation (2nd Edition), page 163-164. There is a certain step in the proof, which I can not fill the details. First, I mention a relevant proposition and a definition.

Proposition: The infinite product $$\prod_{n=1}^{\infty}(1+a_k)$$ converges if $$\sum_{n=1}^{\infty}|a_n|<\infty$$, and in that case $$\left | \prod_{k=1}^\infty (1+a_k)-1 \right |\leq e^{\sum_{k=1}^\infty |a_k|}-1 \tag{*}$$

Definition: Let $$(a_n)_{n\geq 1}$$ be a sequence of complex valued functions defined in an open subset $$\Omega$$ of $$\mathbb{C}$$. We say that the infinite product $$\prod_{n=1}^{\infty}(1+a_k)$$ converges locally uniformly in $$\Omega$$, if $$\prod_{n=1}^{\infty}(1+a_k(z))$$ converges at each $$z\in \Omega$$ and if furthermore to each compact subset $$K$$ of $$\Omega$$ and each $$\epsilon>0$$ there exists an $$N$$ such that for all $$z\in K$$ and $$n\geq N$$ $$\left | \prod_{k=1}^{\infty}(1+a_k(z))-\prod_{k=1}^{n}(1+a_k(z)) \right |<\epsilon.$$

What I want to prove is:

Lemma: Let $$(a_n)_{n\geq 1}$$ be a sequence of complex valued functions on an open subset $$\Omega$$ of $$\mathbb{C}$$. If as $$N\to\infty$$ the sum $$\sum_{n=N}^{\infty}|a_n(z)|$$ converges locally uniformly to $$0$$, then the infinite product $$\prod_{n=1}^{\infty}(1+a_k)$$ converges locally uniformly in $$\Omega$$.

The proof which the author says is simply: "It is a consequence of Proposition and the inequality (*)" It does not seem completely clear to me. Could someone explain that step for me?

Update: When I read the answer of Kavi Rama Murthy, I thought as follows: Let $$K$$ be any compact subset of $$\Omega$$, and let $$\epsilon\in (0,1/2)$$ be given. Choose an integer $$N^*$$ with $$N^*\geq N$$ such that $$\sum_{n=n+1}^{\infty}|a_n(z)|<\epsilon$$ for all $$z\in K$$ and all $$n\geq N^*$$. Then, we have for all $$z\in K$$ and all $$n\geq N^*$$ $$\left | \prod_{k=n+1}^\infty (1+a_k(z))-1 \right |\leq e^{\sum_{k=n+1}^\infty |a_k(z)|}-1 and so $$\left | \prod_{k=1}^{\infty}(1+a_k(z))-\prod_{k=1}^{n}(1+a_k(z)) \right |=\left | \prod_{k=1}^{n}(1+a_k(z)) \right |\left | \prod_{k=n+1}^{\infty}(1+a_k(z))-1 \right |<2\epsilon \left | \prod_{k=1}^{n}(1+a_k(z)) \right |.$$ Then I got stuck here.

• I'd like to help but I don't know what "Proposition and the inequality (*)" is. I don't have the book. Commented Mar 14, 2022 at 1:51
• @DanielWainfleet They are stated in the beginning of the post. I think I have figured out the proof, but thanks for trying to help. If you want and have time, could you please see my other post? Commented Mar 14, 2022 at 5:26

$$|\prod_{k=n}^{m}(1+a_k(z))-1|\leq e^{ \sum\limits_{k=n}^{m}|a_k(z)}-1$$. On any compact set $$K$$ we can choose $$N$$ such that $$e^{ \sum\limits_{k=n}^{m}|a_k(z)}-1<\epsilon$$ for alll $$z \in K$$ fro all $$n,m \geq N$$. Now consider $$|\prod_{k=1}^{n}(1+a_k(z))-\prod_{k=1}^{m}(1+a_k(z))|$$ where $$n . We can write this as $$|\prod_{k=1}^{n}(1+a_k(z))| |\prod_{k=n+1}^{m}(1+a_k(z))-1|$$. What remains is to see that th first factor $$\prod_{k=1}^{n}(1+a_k(z))$$ is uniformly bounded on $$K$$. This follows by another application of (*). I hope you can finish.
• @Mr.MathDoctor $\left | \prod_{k=1}^{n}(1+a_k(z)) \right |\leq 1+(e^{\sum_{k=1}^{n} a_k(z)}-1)$ and $\sum_{k=1}^{n} a_k(z)$ is bounded on $K$, right? Commented Mar 14, 2022 at 0:13
• That was my initial thought, but then I thought it might depend on $z$, wouldn't it? Is it because that the series $\sum_{k=1}^{\infty} |a_k(z)|$ is uniformly convergent on $K$, it must be bounded there? Commented Mar 14, 2022 at 0:18
• @Mr.MathDoctor There exists $N$ such that $|\sum\limits_{k=N}^{\infty} a_k(z)|<1$ for all $z \in K$. Commented Mar 14, 2022 at 0:26
• Ok, that makes sense. I think I should add an extra assumption in the lemma, that the sequence $(a_n)_{n\geq 1}$ is of bounded complex valued functions, otherwise it is not known whether the rest of function terms $a_1(z),..., a_{N-1}$ are bounded or not, yet the sum of all other terms is bounded, right? Commented Mar 14, 2022 at 0:33