Function on $\mathbb{C}$ with all primes as zeros?

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???
• Note that $\sum\frac 1p$ does not converge... – abiessu Jun 7 '20 at 12:38
• Shouldn't this converge for all $z$? It looks like a Cauchy sequence to me. – Alexander Geldhof Jun 7 '20 at 12:38
• Reflect on your point 1. This is why the Weierstrass factorisation theorem needs terms like $e^{z/p}$. – Angina Seng Jun 7 '20 at 12:40
• @abiessu: fixed, the not was missing. – Harald Jun 7 '20 at 12:42
• 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). – Martin R Jun 7 '20 at 12:46

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 $$\prod_{p \text{ is prime}} \left(1-\frac{z}{p}\right)\, .$$