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According to the Weierstraß factorization theorem, an entire function with all primes as zeros would be (if I didn't mess up):

$$\tilde P(z) = \prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right) \cdot e^{z/p} $$ for $z\in\mathbb{C}$.

Formally I can define the slightly different function $$P(z) = \prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right)$$

for $z\in\mathbb{C}$.

Question: For which $z\in\mathbb{C}$ does the latter converge apart from the obvious $z=0$ or $z$ is prime?

Thoughts:

  1. Taking the $\log$ this translates to the question for which $z$ $$\sum_{p \text{ is prime}}\log(1-z/p)$$ converges. I am guessing that this is not the case because thinning the harmonic series out to just primes, $\sum 1/p$, does not make it convergent and since $\sum \log(1-1/n)$ is not convergent, thinning out to primes likely does not help either.
  2. Yet for $\tilde P$ the $\log$ yields $$\sum_{p \text{ is prime}}(z/p+\log(1-z/p)),$$ which is, in absolute value, even larger but should converge because $\tilde P(z)$ converges. Hmm???
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    $\begingroup$ Note that $\sum\frac 1p$ does not converge... $\endgroup$ – abiessu Jun 7 '20 at 12:38
  • $\begingroup$ Shouldn't this converge for all $z$? It looks like a Cauchy sequence to me. $\endgroup$ – Alexander Geldhof Jun 7 '20 at 12:38
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    $\begingroup$ Reflect on your point 1. This is why the Weierstrass factorisation theorem needs terms like $e^{z/p}$. $\endgroup$ – Angina Seng Jun 7 '20 at 12:40
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    $\begingroup$ @abiessu: fixed, the not was missing. $\endgroup$ – Harald Jun 7 '20 at 12:42
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    $\begingroup$ What do you mean with “in absolute value, even larger”? – Note that $z/p+\log(1-z/p) \sim (z/p)^2$ for $p \to \infty$, and that makes the series (2) converge (uniformly on compact sets). $\endgroup$ – Martin R Jun 7 '20 at 12:46
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Remark: There are two slightly different definitions of convergence of an infinite product $\prod_{n=1}^\infty a_n$. One definition is that all factors $a_n$ are nonzero and that $\lim_{N \to \infty} \prod_{n=1}^N$ exists and it not zero.

The other definition allows that finitely many factors are zero, and requires that for some $n_0$, $\lim_{N \to \infty} \prod_{n=n_0}^N$ exists and it not zero. Using this definition, the value of a convergent infinite product is zero if and only if one factor is zero.

With the first definition, both of your products are divergent if $z$ is a prime.

I'll use the second definition here, which has the advantage that the case “$z$ is a prime” does not need to be considered separately.


It follows e.g. from the Taylor series of $\log(1+w)$ that $$ |\log(1+w) - w | < K |w|^2 $$ for $|w| < 1/2$ and some constant $K > 0$. Therefore, for $|z| < R$ and all primes $p > 2|z|$, $$ \tag{*} \left| \log\left(1-\frac{z}{p}\right) + \frac{z}{p}\right| < \frac{KR^2}{p^2} \, . $$ This implies the (locally uniform) convergence of $\sum\limits_{p > 2|z| \text{ is prime}} \log\left(1-\frac{z}{p}\right) + \frac{z}{p}$ and thus the convergence of the infinite product $$\prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right) \cdot e^{z/p}\, .$$

For non-zero $z$ is the series $\sum\limits_{p \text{ is prime}} \frac{z}{p}$ divergent, so that $(*)$ also implies the divergence of $\sum\limits_{p > 2|z| \text{ is prime}} \log\left(1-\frac{z}{p}\right)$ and consequently the divergence of $$\prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right)\, .$$

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