# If a product of polynomials converges, does some product of their zeros also converge?

Suppose $$\{p_k\}$$ is a sequence of polynomials with $$p_k(0)=1$$. Let $$a_1,a_2,\ldots$$ be an enumeration of all of the zeros of the $$p_k$$. Suppose that $$\prod_{k=1}^\infty p_k(z)$$ converges uniformly on compact subsets of $$\mathbb{C}$$. Is there some permutation $$\sigma:\mathbb{N}\to\mathbb{N}$$ such that $$f_\sigma (z)=\prod_{k=1}^\infty \left(1-\frac{z}{a_{\sigma(k)}}\right)$$ converges uniformly on compact subsets of $$\mathbb{C}$$?

This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such $$\{p_k\}$$ is given along with a permutation $$\sigma$$ for which the product $$f_\sigma (z)$$ does $$\underline{\text{not converge}}$$.

## This question had a bounty worth +50 reputation from user122916 that ended 13 hours ago. Grace period ends in 10 hours

This question has not received enough attention.

• Since you have $a_{\sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$. – Robert Israel Dec 6 at 15:56
• Thank you. I meant $p_k(0)=1$. – user122916 Dec 6 at 15:57
• My intuition is telling me that the answer is definitely "no". But I need to think more. – mathworker21 Dec 9 at 3:00
• I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf – Lukas Geyer Dec 11 at 23:45

Let $$p$$ be any prime number.
Now, $$i^{p^s} = 1$$ for $$s\neq 1$$. Now if your statement is correct, we can apply logarithm to both sides and prove that product of prime converges to $$0$$ which is not true.
I have shown you the answer by taking "$$s$$" dimension polynomial, whose roots are all the prime numbers.