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  "body": "\n## Statement\n\nLet $(a_n)_{n \\ge 1}$ be a sequence of positive real numbers.\nThen the infinite product $\\prod_{n=1}^{\\infty} a_n$ [[converges|H1Z2TKY3EQ]] if and only if the series $\\sum_{n=1}^{\\infty} \\ln a_n$ [[converges|0CVVKY7GY2]], where $\\ln$ is the [[natural logarithm|0044CPHJBV]].\nIn that case\n$$\\prod_{n=1}^{\\infty} a_n = \\exp\\left( \\sum_{n=1}^{\\infty} \\ln a_n \\right).$$\n\n## Remarks\n\nThe partial products and the partial sums are linked by $\\prod_{n \\le N} a_n = \\exp\\left(\\sum_{n \\le N} \\ln a_n\\right)$.\nThe map $\\exp$ is a continuous bijection from $\\R$ onto $(0, \\infty)$, with continuous inverse $\\ln$.\nSo the partial products tend to a positive limit exactly when the partial sums tend to a real limit.\nThe nonzero condition in the definition of convergence is what makes this work: partial sums tending to $-\\infty$ correspond to products diverging to $0$.\nA consequence: if $u_n \\ge 0$, then $\\prod (1 + u_n)$ converges if and only if $\\sum u_n$ converges.\nIndeed, if $u_n \\to 0$ then $\\ln(1 + u_n) / u_n \\to 1$, and the [[limit comparison test|0HW7BNHHGF]] applies; if $u_n \\not\\to 0$, both diverge.\nExample: $\\prod (1 + 1/n^2)$ converges and $\\prod (1 + 1/n)$ diverges.\n\n## Explanations\n\n### L1 human\n\nTaking logarithms turns products into sums.\nSo an infinite product of positive numbers behaves exactly like the infinite series of their logarithms.\nThis lets the convergence tests for series decide convergence of products.\n\n### L2 human\n\nThe statement reduces the theory of products of positive numbers to the theory of series.\nIts most used form is the criterion for $\\prod (1 + u_n)$ with $u_n \\ge 0$, which compares the product with $\\sum u_n$.\nFor complex factors one uses a branch of the logarithm near $1$, and absolute convergence of $\\sum (a_n - 1)$ gives convergence of the product.\n",
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