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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.

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  • $\begingroup$ Is the first product assumed to converge pointwise everywhere? And does "convergent" mean convergent to a nonzero value? $\endgroup$ – zhw. Jul 6 '18 at 22:53
  • $\begingroup$ The function $f(z)$ is assumed to be an entire function, so yes, the product converges uniformly on compact subsets of $\mathbb{C}$. $\endgroup$ – user122916 Jul 7 '18 at 4:22
  • $\begingroup$ Just to make sure I understand: if you order the $r_j$'s based on their appearance in the sequence $p_1,p_2,p_3,\dots$, then the fact that $\prod p_k$ converges obviously implies $\prod (1-\frac{z}{r_j})$ converges. But you are asking whether $\prod (1-\frac{z}{r_j})$ converges absolutely, i.e. whether any ordering of the $r_j$'s makes $\prod (1-\frac{z}{r_j})$ converge (conditionally). Is this accurate? $\endgroup$ – mathworker21 Jul 8 '18 at 8:06
  • $\begingroup$ Actually, it is that "obviously implies" part that I am stuck on. Could you explain a bit? $\endgroup$ – user122916 Jul 8 '18 at 14:37
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$\prod_{n=1}^\infty (1-\frac{z^2}{n^2})$ 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.

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  • $\begingroup$ I think the issue is that the degree of the $p_k$'s does not need to be bounded. I think I have a counterexample, and will type it up in an answer below. $\endgroup$ – user122916 Jul 12 '18 at 20:39
  • $\begingroup$ My example doesn't work :). I am trying to turn this argument into an $\epsilon-\delta$ style argument, but I am having issues. In fact, I cannot even convince myself that $\lim_{N\to \infty} \prod_{j=1}^N \left(1-\frac{z}{r_j}\right)$ converges. Any suggestions? I suppose I am looking to more precisely show that the intermediate products "won't really matter." $\endgroup$ – user122916 Jul 12 '18 at 21:38

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