Weierstrass product

Remarks

Since $\abs{1 - E_p(z)} \le \abs{z}^{p+1}$ for $\abs{z} \le 1$, the product converges uniformly on compact subsets of $\C$. So $P$ is an entire function, and its zeros are exactly the $a_n$, with multiplicity the number of indices $n$ with $a_n = a$. The choice $p_n = n - 1$ always satisfies the condition. Example: $a_n = n$ and $p_n = 1$; then $\sum_n (r/n)^2 < \infty$, so $\prod_{n \ge 1} (1 - z/n) e^{z/n}$ is a Weierstrass product, with zeros at the positive integers. Non-example: $a_n = n$ and $p_n = 0$ fail the condition, since $\sum_n r/n$ diverges; indeed $\prod_n (1 - z/n)$ diverges at $z = -1$. When all $p_n$ equal the least integer $p \ge 0$ with $\sum_n \abs{a_n}^{-(p+1)} < \infty$, the product is called the canonical product, and $p$ is its genus.