Alternate Divergent Series of zeta Function

SPECIES

$\displaystyle \sum_{n\geq0} (-1)^n \zeta(n) = \dfrac{1}{2} - \gamma$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \zeta(n) = \gamma - 1$ (convergent)

$\displaystyle \sum_{n\geq2} (-1)^n \zeta(n) = 1$ (convergent)

$\displaystyle \sum_{n\geq0} (-2)^n \zeta(n) = \dfrac{9}{2} - 2\gamma$ (stable)

$\displaystyle \sum_{n\geq1} (-2)^{n-1} \zeta(n) = -\psi^{(0)}(3) = -\dfrac{3}{2} + \gamma$ (stable)

$\displaystyle \sum_{n\geq1} (-2)^n \zeta(n) = 2\psi^{(0)}(3) = 3 - 2\gamma$ (stable)

$\displaystyle \sum_{n\geq2} (-2)^n \zeta(n) = 3$ (stable)

$\displaystyle \sum_{n\geq1} (-3)^{n-1} \zeta(n) = -\psi^{(0)}(4) = -\dfrac{11}{6} + \gamma$ (stable)

$\displaystyle \sum_{n\geq1} (-3)^n \zeta(n) = 3\psi^{(0)}(4) = \dfrac{11}{2} - 3\gamma$ (stable)

$\displaystyle \sum_{n\geq1} (-4)^{n-1} \zeta(n) = -\psi^{(0)}(5) = -\dfrac{25}{12} + \gamma$ (stable)

$\displaystyle \sum_{n\geq1} (-4)^n \zeta(n) = 4\psi^{(0)}(5) = \dfrac{25}{3} - 4\gamma$ (stable)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{n} = 0$ (convergent)

$\displaystyle \sum_{n\geq2} (-1)^n \dfrac{\zeta(n)}{n} = \gamma$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{2^{n-1}} = -2 + 2\ln2 + \gamma$ (convergent)

$\displaystyle \sum_{n\geq0} (-1)^n \dfrac{\zeta(n)}{2^n} = \dfrac{1}{2} - \ln2 - \dfrac{\gamma}{2}$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{2^n} = -1 + \ln2 + \dfrac{\gamma}{2}$ (convergent)

$\displaystyle \sum_{n\geq2} (-1)^n \dfrac{\zeta(n)}{2^n} = 1 - \ln2$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{3^{n-1}} = -3 + \dfrac{\pi}{2\sqrt{3}} + \dfrac{3\ln3}{2} + \gamma$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{3^n} = -1 + \dfrac{\pi}{6\sqrt{3}} + \dfrac{\ln3}{2} + \dfrac{\gamma}{3}$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{4^{n-1}} = -4 + \dfrac{\pi}{2} + 3\ln2 + \gamma$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{4^n} = -1 + \dfrac{\pi}{8} + \dfrac{3\ln2}{4} + \dfrac{\gamma}{4}$ (convergent)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{2^n\,\zeta(n)}{n} = -\ln\Gamma(3) = -\ln 2$ (stable)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{3^n\,\zeta(n)}{n} = -\ln\Gamma(4) = -\ln 6$ (stable)

$\displaystyle \sum_{n\geq1} (-1)^{n-1} \dfrac{\zeta(n)}{n\,2^n} = \ln2 - \dfrac{\ln \pi}{2}$ (convergent)

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