Divergent Products of Rational Numbers

HYPERSECTIONS

$\displaystyle \prod_{n\geq1} (n\!+\!b)^{\frac{1+(-1)^{n-1}}{2}} = (1\!+\!b)\,.\,1\,.\,(3\!+\!b)\,.\,1\,.\,(5\!+\!b)\,.\,1\,.\,\ldots \\ = 2^{\frac{1+b}{2}} \; \dfrac{\Gamma(1\!+\!\frac{b}{2})}{\Gamma(1\!+\!b)}$ (unstable)

$\displaystyle \prod_{n\geq1} (n\!+\!b)^{\frac{1-(-1)^{n-1}}{2}} = 1\,.\,(2\!+\!b)\,.\,1\,.\,(4\!+\!b)\,.\,1\,.\,(6\!+\!b)\,.\,1\,.\,\ldots \\ = \dfrac{\sqrt{\pi}}{2^{\frac{b}{2}}\,\Gamma(1\!+\!\frac{b}{2})}$ (unstable)

SECTIONS

$\displaystyle \prod_{n\geq1} a^{(-1)^{n-1}} = a\,.\,\dfrac{1}{a}\,.\,a\,.\,\dfrac{1}{a}\,.\,a\,.\,\dfrac{1}{a}\,.\,a\,.\,\dfrac{1}{a}\,.\,\ldots = \sqrt{a}$ (semi-stable)

$\displaystyle \prod_{n\geq1} a^{(-1)^n} = \dfrac{1}{a}\,.\,a\,.\,\dfrac{1}{a}\,.\,a\,.\,\dfrac{1}{a}\,.\,a\,.\,\dfrac{1}{a}\,.\,a\,.\,\ldots = \dfrac{1}{\sqrt{a}}$ (semi-stable)

$\displaystyle \prod_{n\geq1} (an)^{(-1)^{n-1}} = a\,.\,\dfrac{1}{2a}\,.\,3a\,.\,\dfrac{1}{4a}\,.\,5a\,.\,\dfrac{1}{6a}\,.\,7a\,.\,\dfrac{1}{8a}\,.\,\ldots = \dfrac{\sqrt{2a}}{\sqrt{\pi}}$ (unstable)

$\displaystyle \prod_{n\geq1} (an)^{(-1)^n} = \dfrac{1}{a}\,.\,2a\,.\,\dfrac{1}{3a}\,.\,4a\,.\,\dfrac{1}{5a}\,.\,6a\,.\,\dfrac{1}{7a}\,.\,8a\,.\,\ldots = \dfrac{\sqrt{\pi}}{\sqrt{2a}}$ (unstable)

$\displaystyle \prod_{n\geq1} (n\!+\!b)^{(-1)^{n-1}} = (1\!+\!b)\,.\,\dfrac{1}{2\!+\!b}\,.\,(3\!+\!b)\,.\,\dfrac{1}{4\!+\!b}\,.\,(5\!+\!b)\,.\,\dfrac{1}{6\!+\!b}\,.\,\ldots \\ = \dfrac{\sqrt{2}}{\sqrt{\pi}} \; 2^b \; \dfrac{\Gamma^2(1\!+\!\frac{b}{2})}{\Gamma(1\!+\!b)}$ (unstable)

$\displaystyle \prod_{n\geq1} (n\!+\!b)^{(-1)^n} = \dfrac{1}{1\!+\!b}\,.\,(2\!+\!b)\,.\,\dfrac{1}{3\!+\!b}\,.\,(4\!+\!b)\,.\,\dfrac{1}{5\!+\!b}\,.\,(6\!+\!b)\,.\,\ldots \\ = \dfrac{\sqrt{\pi}}{\sqrt{2}} \; 2^{-b} \; \dfrac{\Gamma(1\!+\!b)}{\Gamma^2(1\!+\!\frac{b}{2})}$ (unstable)

$\displaystyle \prod_{n\geq1} (an\!+\!b)^{(-1)^{n-1}} = (a\!+\!b)\,.\,\dfrac{1}{2a\!+\!b}\,.\,(3a\!+\!b)\,.\,\dfrac{1}{4a\!+\!b}\,.\,(5a\!+\!b)\,.\,\ldots \\ = \dfrac{\sqrt{2a}}{\sqrt{\pi}} \; 2^{\frac{b}{a}} \; \dfrac{\Gamma^2(1\!+\!\frac{b}{2a})}{\Gamma(1\!+\!\frac{b}{a})}$ (unstable)

$\displaystyle \prod_{n\geq1} (an\!+\!b)^{(-1)^n} = \dfrac{1}{a\!+\!b}\,.\,(2a\!+\!b)\,.\,\dfrac{1}{3a\!+\!b}\,.\,(4a\!+\!b)\,.\,\dfrac{1}{5a\!+\!b}\,.\,\ldots \\ = \dfrac{\sqrt{\pi}}{\sqrt{2a}} \; 2^{-\frac{b}{a}} \; \dfrac{\Gamma(1\!+\!\frac{b}{a})}{\Gamma^2(1\!+\!\frac{b}{2a})}$ (unstable)

Euler’s reflection formula

$\displaystyle \prod_{n\geq1} \dfrac{n^2}{n^2\!-\!a^2} = \dfrac{1}{1\!-\!a^2}\cdot\dfrac{4}{4\!-\!a^2}\cdot\dfrac{9}{9\!-\!a^2}\cdot\dfrac{16}{16\!-\!a^2}\cdot\,\cdots \\ = \Gamma(1\!+\!a)\,\Gamma(1\!-\!a) = \dfrac{\pi a}{\sin(\pi a)}$ (convergent)

$\displaystyle \prod_{n\geq1} \dfrac{n^2\!-\!a^2}{n^2} = (1\!-\!a^2)\cdot\dfrac{4\!-\!a^2}{4}\cdot\dfrac{9\!-\!a^2}{9}\cdot\dfrac{16\!-\!a^2}{16}\cdot\,\cdots \\ = \dfrac{1}{\Gamma(1\!+\!a)\,\Gamma(1\!-\!a)} = \dfrac{\sin(\pi a)}{\pi a}$ (convergent)

SUBSECTIONS

Euler’s truncated reflection formula

$\displaystyle \prod_{n\geq2} \dfrac{n^2}{n^2-a^2} = \prod_{n\geq1} \dfrac{(n+1)^2}{(n+1)^2-a^2} = \dfrac{4}{4-a^2}\cdot\dfrac{9}{9-a^2}\cdot\dfrac{16}{16-a^2}\cdot\,\cdots \\ = \Gamma(2+a)\,\Gamma(2-a) = \dfrac{\pi a}{\sin(\pi a)}\cdot(1-a^2)$ (convergent)

$\displaystyle \prod_{n\geq3} \dfrac{n^2}{n^2-a^2} = \prod_{n\geq1} \dfrac{(n+2)^2}{(n+2)^2-a^2} = \dfrac{9}{9-a^2}\cdot\dfrac{16}{16-a^2}\cdot\dfrac{25}{25-a^2}\cdot\,\cdots \\ = \dfrac{\Gamma(3+a)\,\Gamma(3-a)}{4} = \dfrac{\pi a}{\sin(\pi a)}\cdot\dfrac{(1-a^2)\,(4-a^2)}{4}$ (convergent)

$\displaystyle \prod_{n\geq4} \dfrac{n^2}{n^2-a^2} = \prod_{n\geq1} \dfrac{(n+3)^2}{(n+3)^2-a^2} = \dfrac{16}{16-a^2}\cdot\dfrac{25}{25-a^2}\cdot\dfrac{36}{36-a^2}\cdot\,\cdots \\ = \dfrac{\Gamma(4+a)\,\Gamma(4-a)}{36} = \dfrac{\pi a}{\sin(\pi a)}\cdot\dfrac{(1-a^2)\,(4-a^2)\,(9-a^2)}{36}$ (convergent)

INFRASECTIONS

Euler’s truncated reflection formula (generalized)

$\displaystyle \prod_{n\geq m+1} \dfrac{n^2}{n^2-a^2} = \prod_{n\geq1} \dfrac{(n+m)^2}{(n+m)^2-a^2} \\ = \dfrac{(m+1)^2}{(m+1)^2-a^2}\cdot\dfrac{(m+2)^2}{(m+2)^2-a^2}\cdot\dfrac{(m+3)^2}{(m+3)^2-a^2}\cdot\,\cdots \\ = \dfrac{\Gamma(m+1+a)\,\Gamma(m+1-a)}{\prod_{k=1}^m k^2} = \dfrac{\pi a}{\sin(\pi a)}\cdot\prod_{k=1}^m\left(1-\left(\frac{a}{k}\right)^2\right)$ (convergent)

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