Does there exist $z_0 \notin \{a_n\}_{n \geq 1}$ such that $f(z_0) = 0\ $? I am reading Weierstrass Factorization Theorem from the lecture notes given by our instructor. Here I came across a theorem which is the following $:$

$\mathbf {Theorem} :$ Let $\{a_n\}_{n \geq 1}$ be a sequence of non-zero complex numbers such that $\lim\limits_{n \to \infty} |a_n| = \infty.$ If $\{p_n\}_{n \geq 1}$ is a sequence of positive integers such that $$\sum\limits_{n = 1}^{\infty} \left (\frac {r} {|a_n|} \right )^{p_n + 1} \lt \infty$$ for all $r \gt 0$ then $$f(z) = \prod\limits_{n=1}^{\infty} E_{p_n} (z/a_n)$$ converges in $H(\mathbb C)$ to an entire function $f.$

Here $E_p$'s are the elementary factor defined for $p =0,1,2,\cdots$ as follows $:$
$$\begin{align*} E_0 (z) & = 1 - z, \\ E_p(z) & = (1 - z) \exp \left (z + \frac {z^2} {2} + \cdots + \frac {z^p} {p} \right ),\ p \geq 1. \end{align*}$$
From here our instructor concluded that $f$ has zeros only at the points $a_n.$
I understand that $f$ has zeros at $a_n$ since $E_{p_n} (z/a_n)$ has a simple zero at $a_n$ and there are no other zeros. But I don't understand the fact that $f$ can't have any other zero. The infinite product is something which confuses me. Because if we have finitely many non-zero quantities their product is still non-zero but this is not true for infinitely many non-zero quantities. For instance if we take $b_n = \frac {1} {2},$ for all $n \geq 1$ then their product is clearly $0$ if we don't restrict ourselves to finitely many $n.$ Could anyone help me understanding this concept?
Any help in this regard would be warmly appreciated. Thanks for investing your valuable time in reading my question.
 A: To answer your question, let's recall a few facts about infinite products.

*

*Suppose $a_n\neq-1$ for all $n\in\mathbb{N}$ and define $q_n=\prod^n_{k=1}(1+a_k)$.


Definition: $\prod_n(1+a_n)$ converges if there is  $p\neq0$  such that $\lim_{n\rightarrow\infty}q_n=p$. When $\lim_{n\rightarrow\infty}q_n=0$, it is said that $\prod_n(1+a_n)$ diverges to $0$.

It is a well known fact (Cauchy criteria for infinite products) that $\prod_n(1+a_n)$ converges iff for any $\varepsilon>0$ there is $N_\varepsilon>0$ such that
$$\Big|\prod^k_{j=1}(1+a_{n+k}) -1\Big|<\varepsilon,\qquad n\geq N, \, k\in\mathbb{N}$$
See for example, Apostol, T.,  Mathematical Analysis, p.p. 207. The proof is based on the standard Cauchy criteria of sequences. If you need details about his, let me know and will provide some steps for you to complete the proof.


*Recall that a product $q_n=\prod^n_{k=1}(1+a_n)$ is said to converge absolutely (again $a_n\neq-1$ for all $n\in\mathbb{N}$) iff $Q_n=\prod^n_{k=1}(1+|a_k|)$ converges. Form the Cauchy criteria above,  it is not difficult to check that if $q_n$ converges absolutely, then $p=\lim_nq_n$ exists and $p\neq0$.


*Finally, recall that $q_n=\prod^n_{k=1}(1+a_n)$  converges absolutely (again $a_n\neq-1$ for all $n\in\mathbb{N}$) absolutely iff $\sum_n|a_n|<\infty$. This follows from the inequality
$$ \sum^n_{k=1}|a_k|\leq \prod^n_{k=1}(1+|a_k|)\leq \exp\Big({\sum^n_{k=1}|a_n|}\Big)$$

In the product $\prod^\infty_{n=1}E_{p_k}(z/a_n)$ where $|a_n|\xrightarrow{n\rightarrow\infty}\infty$ you have the estimates
$$|1-E_{p_n}(z/a_n)|\leq |z|^{p_n+1}\frac{1}{|a_n|^{p_n+1}}$$
for $|z|<|a_n|$, with $$\sum^\infty_{n=1}\Big(\frac{r}{|a_n|}\Big)^{k+1}<\infty$$ for every fixed $r>0$.  This shows that the product $\prod_nE_n(z/a_n)=\prod_n \big(1-(1-E_k(z/a_n))\big)$ converges absolutely when $E_k(z/a_n)\neq0$ for all $n$, and vanishes only if $z=a_n$ for some $n$.
