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Question Number 193204 by Mingma last updated on 07/Jun/23

Answered by witcher3 last updated on 08/Jun/23

$${\mathrm{H}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}}={\stackrel{1}{\int }}_0\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{\mathrm{x}}^{\mathrm{k}-1}\mathrm{dx}$$$$={\int }_0^1\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}\mathrm{dx}$$$$=\underset{\mathrm{a}\to 1}{\lim }{\int }_0^{\mathrm{a}}\frac{1-{\mathrm{x}}^{\mathrm{n}}}{1-\mathrm{x}}\mathrm{dx}=\underset{\mathrm{a}\to 1}{\lim }[{}_0^{\mathrm{a}}-\ln (1-\mathrm{x})(1-{\mathrm{x}}^{\mathrm{n}})]$$$$-\int {\mathrm{nx}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$$$\iff {\mathrm{H}}_{\mathrm{n}}=-\mathrm{n}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})$$$$\iff \frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^3}=-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}-1}}{{\mathrm{n}}^3}\ln (1-\mathrm{x})\mathrm{dx}$$$$\Rightarrow \sum _{\mathrm{n}⩾1}\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^3}=-{\int }_0^1\frac{1}{\mathrm{x}}\left(\sum _{\mathrm{n}⩾1}\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2}\right)\ln (1-\mathrm{x})\mathrm{dx}$$$$=-{\int }_0^1\frac{\ln (1-\mathrm{x})}{\mathrm{x}}{\mathrm{Li}}_2\left(\mathrm{x}\right)\mathrm{dx}$$$${\mathrm{Li}}_2\left(\mathrm{z}\right)=-{\int }_0^{\mathrm{z}}\frac{\ln (1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\iff \sum _{\mathrm{n}⩾1}\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^3}=\frac{1}{2}{\mathrm{Li}}_2^2\left(1\right)=\frac{1}{2}{\left(\frac{{\pi }^2}{6}\right)}^2=\frac{{\pi }^4}{72}$$$$$$$$$$

Commented by Mingma last updated on 09/Jun/23

Perfect ��

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