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  "body": "\n## Definition\n\nFor an integer $p \\ge 0$, the *elementary factor* $E_p$ is the entire function\n$$E_0(z) = 1 - z, \\qquad E_p(z) = (1 - z) \\exp\\left(\\sum_{k=1}^{p} \\frac{z^k}{k}\\right) \\quad (p \\ge 1).$$\nLet $(a_n)_{n \\ge 1}$ be a sequence of nonzero complex numbers with $\\abs{a_n} \\to \\infty$, and let $(p_n)_{n \\ge 1}$ be integers $p_n \\ge 0$ such that\n$$\\sum_{n \\ge 1} \\left(\\frac{r}{\\abs{a_n}}\\right)^{p_n + 1} \u003c \\infty \\quad \\text{for every } r \u003e 0.$$\nThe *Weierstrass product* with zeros $(a_n)$ and indices $(p_n)$ is the function\n$$P(z) = \\prod_{n \\ge 1} E_{p_n}\\!\\left(\\frac{z}{a_n}\\right).$$\n\n## Remarks\n\nSince $\\abs{1 - E_p(z)} \\le \\abs{z}^{p+1}$ for $\\abs{z} \\le 1$, the product converges [[uniformly|08FPABDQ4C]] on compact subsets of $\\C$.\nSo $P$ is an [[entire function|0419794R0T]], and its [[zeros|0CEGEASY49]] are exactly the $a_n$, with multiplicity the number of indices $n$ with $a_n = a$.\nThe choice $p_n = n - 1$ always satisfies the condition.\nExample: $a_n = n$ and $p_n = 1$; then $\\sum_n (r/n)^2 \u003c \\infty$, so $\\prod_{n \\ge 1} (1 - z/n) e^{z/n}$ is a Weierstrass product, with zeros at the positive integers.\nNon-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 $z = -1$.\nWhen all $p_n$ equal the least integer $p \\ge 0$ with $\\sum_n \\abs{a_n}^{-(p+1)} \u003c \\infty$, the product is called the canonical product, and $p$ is its genus.\n",
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