Factoring a convergent infinite product of polynomials. Suppose that $f(z)=\displaystyle\prod_{k=1}^\infty p_k(z)$ is a convergent product of polynomials $p_k$ such that $p_k(0)=1$. I want to know if I can "factor" $f(z)$ in the following way: if we list the roots of all the $p_k$ as $r_1, r_2, r_3, \ldots$, must the product $\displaystyle\prod_{j=1}^\infty \left( 1-\frac{z}{r_j}\right)$ converge?
I understand that the Weierstrass Factorization Theorem gives a factorization for $f(z)$ that involves exponential terms to ensure convergence, but I am wondering whether knowing only that $\displaystyle\prod_{k=1}^\infty p_k(z)$ converges is enough to conclude that the roots grow fast enough.

$\prod_{n=1}^\infty (1-\frac{z^2}{n^2})$ converges locally uniformly on $\mathbb{C}$. We can let $p_1 = \prod_{n=1}^{N_1} (1-\frac{z^2}{n^2}), p_2 = \prod_{n=N_1+1}^{N_2} (1-\frac{z^2}{n^2}), p_3 = \prod_{n=N_2+1}^{N_3} (1-\frac{z^2}{n^2})$, etc. for positive integers $N_1 < N_2 < \dots$. We may order the roots of $p_1$ as $-1,-2,...,-N_1,1,2,\dots,N_1$, and order the roots of $p_2$ as $-(N_1+1),\dots,-N_2, (N_1+1),\dots, N_2$, etc. The point is that $\prod_{n=1}^\infty (1+\frac{z}{n})$ goes to infinity if $z \in \mathbb{R}^+$, so if $z \in \mathbb{R}^-$, then $\prod_{n=N_j+1}^{N_{j+1}} (1+\frac{z}{-n})$ will be large for $z \in \mathbb{R}^-$. So we can choose the $N_i$'s to be spaced out enough so that we don't get local uniform convergence. Note exactly what's going on is that at the cutoff of each $p_k$, we are fine, since we multiplied together the $(1+\frac{z}{n})$s with the $(1-\frac{z}{n})$s, but the intermediate product (exactly "half way" between $p_k$ and $p_{k+1}$) is the issue.
