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Question Number 95784 by abdomathmax last updated on 27/May/20

$$\mathrm{calculate}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}$$$${\mathrm{H}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}$$

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