Aperiodic Monotonic Divergent Series of Rational Numbers

SPECIES

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

$\displaystyle \sum_{n\geq0} \dfrac{1}{n} = \displaystyle \sum_{n\geq1} \dfrac{1}{n-1} = \zeta(1) - \dfrac{1}{2} = \gamma - \dfrac{1}{2}$ (semi-stable)

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

$\displaystyle \sum_{n\geq0} \dfrac{1}{2n} = \dfrac{\gamma}{2} - \dfrac{1}{4}$ (semi-stable)

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

$\displaystyle \sum_{n\geq0} \dfrac{1}{2n+1} = \sum_{n\geq1} \dfrac{1}{2n-1} = \dfrac{1}{2} \big(\gamma-H_{-\frac{1}{2}}\big) = \dfrac{\gamma}{2} + \ln2$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{3n+1} = \sum_{n\geq1} \dfrac{1}{3n-2} =\dfrac{1}{3} \big(\gamma-H_{-\frac{2}{3}}\big) = \dfrac{\gamma}{3} + \dfrac{\pi}{6\sqrt{3}} + \dfrac{\ln3}{2}$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{3n+2} = \sum_{n\geq1} \dfrac{1}{3n-1} = \dfrac{1}{3} \big(\gamma-H_{-\frac{1}{3}}\big) = \dfrac{\gamma}{3} - \dfrac{\pi}{6\sqrt{3}} + \dfrac{\ln3}{2}$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{4n+1} = \sum_{n\geq1} \dfrac{1}{4n-3} = \dfrac{1}{4} \big(\gamma-H_{-\frac{3}{4}}\big) = \dfrac{\gamma}{4} + \dfrac{\pi}{8} + \dfrac{3\ln2}{4}$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{4n+3} = \sum_{n\geq1} \dfrac{1}{4n-1} = \dfrac{1}{4} \big(\gamma-H_{-\frac{1}{4}}\big) = \dfrac{\gamma}{4} - \dfrac{\pi}{8} + \dfrac{3\ln2}{4}$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+1} = \displaystyle \sum_{n\geq1} \dfrac{1}{5n-4} = \dfrac{1}{5} \big(\gamma-H_{-\frac{4}{5}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+1} = \dfrac{\gamma}{5} + \dfrac{\pi}{10}\,\sqrt{1+\dfrac{2}{\sqrt{5}}} + \dfrac{1}{4\sqrt{5}}\,\ln\dfrac{3+\sqrt{5}}{2} + \dfrac{\ln5}{4}$

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+2} = \displaystyle \sum_{n\geq1} \dfrac{1}{5n-3} = \dfrac{1}{5} \big(\gamma-H_{-\frac{3}{5}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+2} = \dfrac{\gamma}{5} + \dfrac{\pi}{10}\,\sqrt{1-\dfrac{2}{\sqrt{5}}} - \dfrac{1}{4\sqrt{5}}\,\ln\dfrac{3+\sqrt{5}}{2} + \dfrac{\ln5}{4}$

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+3} = \displaystyle \sum_{n\geq1} \dfrac{1}{5n-2} = \dfrac{1}{5} \big(\gamma-H_{-\frac{2}{5}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+3} = \dfrac{\gamma}{5} - \dfrac{\pi}{10}\,\sqrt{1-\dfrac{2}{\sqrt{5}}} - \dfrac{1}{4\sqrt{5}}\,\ln\dfrac{3+\sqrt{5}}{2} + \dfrac{\ln5}{4}$

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+4} = \sum_{n\geq1} \dfrac{1}{5n-1} = \dfrac{1}{5} \big(\gamma-H_{-\frac{1}{5}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{5n+4} = \dfrac{\gamma}{5} - \dfrac{\pi}{10}\,\sqrt{1+\dfrac{2}{\sqrt{5}}} + \dfrac{\ln5}{4} + \dfrac{1}{4\sqrt{5}}\,\ln\dfrac{3+\sqrt{5}}{2}$

$\displaystyle \sum_{n\geq0} \dfrac{1}{6n+1} = \displaystyle \sum_{n\geq1} \dfrac{1}{6n-5} = \dfrac{1}{6} \big(\gamma-H_{-\frac{5}{6}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{6n+1} = \dfrac{\gamma}{6} + \dfrac{\pi}{2\sqrt{3}} + \dfrac{\ln2}{3} + \dfrac{\ln3}{4}$

$\displaystyle \sum_{n\geq0} \dfrac{1}{6n+5} = \displaystyle \sum_{n\geq1} \dfrac{1}{6n-1} = \dfrac{1}{6} \big(\gamma-H_{-\frac{1}{6}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{6n+5} = \dfrac{\gamma}{6} - \dfrac{\pi}{2\sqrt{3}} + \dfrac{\ln2}{3} + \dfrac{\ln3}{4}$

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+1} = \displaystyle \sum_{n\geq1} \dfrac{1}{8n-7} = \dfrac{1}{8} \big(\gamma-H_{-\frac{7}{8}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+1} = \dfrac{\gamma}{8} + \dfrac{\pi}{16}\,(\sqrt{2}+1) + \dfrac{\ln2}{2} + \dfrac{1}{8\sqrt{2}}\,\ln(3+\sqrt{2})$

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+3} = \displaystyle \sum_{n\geq1} \dfrac{1}{8n-5} = \dfrac{1}{8} \big(\gamma-H_{-\frac{5}{8}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+3} = \dfrac{\gamma}{8} + \dfrac{\pi}{16}\,(\sqrt{2}-1) + \dfrac{\ln2}{2} - \dfrac{1}{8\sqrt{2}}\,\ln(3+\sqrt{2})$

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+5} = \displaystyle \sum_{n\geq1} \dfrac{1}{8n-3} = \dfrac{1}{8} \big(\gamma-H_{-\frac{3}{8}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+5} = \dfrac{\gamma}{8} - \dfrac{\pi}{16}\,(\sqrt{2}-1) + \dfrac{\ln2}{2} - \dfrac{1}{8\sqrt{2}}\,\ln(3+\sqrt{2})$

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+7} = \displaystyle \sum_{n\geq1} \dfrac{1}{8n-1} = \dfrac{1}{8} \big(\gamma-H_{-\frac{1}{8}}\big)$ (semi-stable)

$\displaystyle \sum_{n\geq0} \dfrac{1}{8n+7} = \dfrac{\gamma}{8} - \dfrac{\pi}{16}\,(\sqrt{2}+1) + \dfrac{\ln2}{2} + \dfrac{1}{8\sqrt{2}}\,\ln(3+\sqrt{2})$

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

Erratum: above sum corrected by Denis Rogov on 14 Oct 2023

$\displaystyle \sum_{n\geq0} \dfrac{1}{n(n+1)} = -\dfrac{1}{2}$ (semi-stable)

$\displaystyle \sum_{n\geq1} \dfrac{1}{n(n+1)} = 1$ (convergent)

$\displaystyle \sum_{n\geq0} \dfrac{1}{n(n-1)} = -1$ (semi-stable)

$\displaystyle \sum_{n\geq1} \dfrac{1}{n(n-1)} = -\dfrac{1}{2}$ (semi-stable)

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

$\displaystyle \sum_{n\geq1} \dfrac{1}{n(n+\frac{1}{2})} = 4(1-\ln2)$ (convergent)

$\displaystyle \sum_{n\geq1} \dfrac{1}{n(n-\frac{1}{2})} = 4 \ln2$ (convergent)

$\displaystyle \sum_{n\geq0} \dfrac{1}{n^2+1} = \dfrac{\pi\coth(\pi)+1}{2}$ (convergent)

$\displaystyle \sum_{n\geq0} \dfrac{1}{n^2-1} = -\dfrac{3}{4}$ (semi-stable)

$\displaystyle \sum_{n\geq1} \dfrac{1}{n^2-1} = \dfrac{1}{4}$ (semi-stable)

$\displaystyle \sum_{n\geq2} \dfrac{1}{n^2-1} = \dfrac{3}{4}$ (convergent)

$\displaystyle \sum_{n\geq1} \dfrac{n}{(n+1)(2n+1)} = \dfrac{\gamma}{2} - \ln2$ (semi-stable)

$\displaystyle \sum_{n\geq1} \dfrac{n}{(2n+1)(3n+1)} = \dfrac{\gamma}{6} - \dfrac{\pi}{6\sqrt{3}} + \ln2 - \dfrac{\ln3}{2}$ (semi-stable)

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