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Question Number 148020 by qaz last updated on 25/Jul/21

$$\underset{\mathrm{k}=2}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{k}}\cdot \frac{\mathrm{lnk}}{\mathrm{k}}=\gamma \ln 2-\frac{1}{2}{\ln }^{2}2$$

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