# Convergence of infinite product of cyclotomic polynomials

From dabbling with p-adics, I remember mentions of $$\infty$$ being the 'infinite prime'. So hence something like: $$\frac{x^\infty -1}{x - 1} = \displaystyle \prod_{\substack{d| p_\infty \\ d \neq 1}} \Phi_{d}(x) = \Phi_{p_\infty}(x) = \sum_{n=0}^{\infty -1} x^n$$ would perhaps make some abstract sense in some formal power series something-or-other. However, if we instead claim that $$\substack{n | \infty \\ n \in \mathbb{N}}$$ we get $$\frac{x^\infty -1}{x - 1} = \displaystyle \prod_{\substack{n=2}}^\infty \Phi_{n}(x)$$, which would be appealing (though to my knowledge, simply untrue) if we work with p-adics etc. For reals, with for instance $$x = \frac{1}{100} \in \mathbb{R}$$, the RHS seems to hover around the 'expected asymptote' value of $$\frac{100}{99}$$ with increasing variance as $$n$$ increases. For the 2-adics, setting $$x=2$$ and taking the logarithm gives an infinite sum of infinite sums, with each $$\log(\Phi_{n}(x))$$ producing a sum starting with a 2-adic valuation given by the first non-zero power of $$x$$ in that polynomial and thence growing, and there are infinitely many of them with $$x^1$$ (e.g. all primes), so the sums would not converge (unless there could be cancellation in some space? dunno).

So after all these incoherent babble, I was wondering what is known about the convergence of the product of all cyclotomic polynomials with or without $$\Phi_1(x)$$ -- and whether the first or second representation of $$\frac{x^\infty -1}{x - 1}$$ make any sense.

Thanks!

• $$\prod_{m=0}^\infty \Phi_{2^m}(x) =\lim_{n \to \infty} \prod_{m=0}^n \Phi_{2^m}(x) = \lim_{n \to \infty} x^{2^n}-1 = -1 \qquad \text{for} \ |x|< 1$$ Jun 11, 2019 at 20:13

The infinite products $$\prod_{n=1}^\infty \Phi_n(x)$$ and $$\prod_{n=2}^\infty \Phi_n(x)$$ diverge for all $$x \ne 0$$. In order to converge, you would certainly need $$\Phi_n(x) \to 1$$ as $$n \to \infty$$, but if $$n$$ is prime $$\Phi_n(x) = 1 + x + x^2 + \ldots + x^{n-1} = 1 + x + O(x^2)$$.