Monotonic Divergent Series of zeta Function

SPECIES

$\displaystyle \sum_{n\geq0} (\zeta(n)-1) = \gamma - \dfrac{3}{2}$ (convergent)

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

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

$\displaystyle \sum_{n\geq0} \left(\zeta(2n)-1\right) = -\dfrac{3}{4}$ (convergent)

$\displaystyle \sum_{n\geq1} \left(\zeta(2n)-1\right) = \dfrac{3}{4}$ (convergent)

$\displaystyle \sum_{n\geq2} \left(\zeta(2n)-1\right) = \dfrac{3}{4} - \dfrac{\pi^2}{6}$ (convergent)

$\displaystyle \sum_{n\geq0} \left(\zeta(2n+1)-1\right) = \gamma - \dfrac{3}{4}$ (convergent)

$\displaystyle \sum_{n\geq1} \left(\zeta(2n+1)-1\right) = \dfrac{1}{4}$ (convergent)

$\displaystyle \sum_{n\geq2} \left(\zeta(2n+1)-1\right) = \dfrac{1}{4} - \zeta(3)$ (convergent)

$\displaystyle \sum_{n\geq0} \zeta(n) = \gamma - 2$ (unstable)

$\displaystyle \sum_{n\geq1} \zeta(n) = \gamma - \dfrac{1}{2}$ (unstable)

$\displaystyle \sum_{n\geq2} \zeta(n) = \dfrac{1}{2}$ (unstable)

$\displaystyle \sum_{n\geq0} \zeta(2n) = -\dfrac{5}{4}$ (unstable)

$\displaystyle \sum_{n\geq1} \zeta(2n) = \dfrac{1}{4}$ (unstable)

$\displaystyle \sum_{n\geq2} \zeta(2n) = \dfrac{1}{4} - \dfrac{\pi^2}{6}$ (unstable)

$\displaystyle \sum_{n\geq0} \zeta(2n+1) = \gamma - \dfrac{5}{4}$ (unstable)

$\displaystyle \sum_{n\geq1} \zeta(2n+1) = -\dfrac{1}{4}$ (unstable)

$\displaystyle \sum_{n\geq2} \zeta(2n+1) = -\dfrac{1}{4} - \zeta(3)$ (unstable)

$\displaystyle \sum_{n\geq0} \left(\zeta(2n)-\zeta(2n+1)\right) = -\gamma$ (semi-stable)

$\displaystyle \sum_{n\geq1} \left(\zeta(2n)-\zeta(2n+1)\right) = \dfrac{1}{2}$ (convergent)

$\displaystyle \sum_{n\geq2} \left(\zeta(2n)-\zeta(2n+1)\right) = \dfrac{1}{2} - \dfrac{\pi^2}{6} + \zeta(3)$ (convergent)

$\displaystyle \sum_{n\geq0} \left(\zeta(2n)+\zeta(2n+1)\right) = \gamma - \dfrac{5}{2}$ (unstable)

$\displaystyle \sum_{n\geq1} \left(\zeta(2n)+\zeta(2n+1)\right) = 0$ (unstable)

$\displaystyle \sum_{n\geq2} \left(\zeta(2n)+\zeta(2n+1)\right) = - \dfrac{\pi^2}{6} - \zeta(3)$ (unstable)

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

$\displaystyle \sum_{n\geq1} 2^n \zeta(n) = 2\gamma - \dfrac{5}{2}$ (unstable)

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

$\displaystyle \sum_{n\geq1} 3^{n-1} \zeta(n) = \gamma - 2$ (unstable)

$\displaystyle \sum_{n\geq1} 3^n \zeta(n) = 3\gamma - 6$ (unstable)

$\displaystyle \sum_{n\geq1} 4^{n-1} \zeta(n) = \gamma - \dfrac{7}{3}$ (unstable)

$\displaystyle \sum_{n\geq1} 4^n \zeta(n) = 4\gamma - \dfrac{28}{3}$ (unstable)

$\displaystyle \sum_{n\geq1} \dfrac{\zeta(n)}{n} = \gamma$ (semi-stable)

$\displaystyle \sum_{n\geq2} \dfrac{\zeta(n)}{n} = 0$ (semi-stable)

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \sum_{n\geq1} \dfrac{\zeta(n)}{4^n} = -\dfrac{\pi}{8} + \dfrac{3\ln 2}{4} + \dfrac{\gamma}{4}$ (convergent)

$\displaystyle \sum_{n\geq0} \dfrac{\zeta(2n)}{2^{2n}} = 0$ (convergent)

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

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

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

$\displaystyle \sum_{n\geq1} \dfrac{\zeta(n)}{n2^n} = \dfrac{\ln\pi}{2}$ (convergent)

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