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Question Number 172785 by mnjuly1970 last updated on 01/Jul/22

Answered by Eulerian last updated on 01/Jul/22

$$$$$$\mathrm{Note}:{\mathrm{H}}_{\mathrm{n}-1}=\gamma +{\psi }^{\left(0\right)}\left(\mathrm{n}\right)$$$${\mathrm{H}}_{\mathrm{n}-1}={\mathrm{H}}_{\mathrm{n}}-\frac{1}{\mathrm{n}}$$$$\zeta \left(\mathrm{z}\right)=\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{k}}^{\mathrm{z}}}\Rightarrow \zeta \left(2\right)=\frac{{\pi }^2}{6}$$$$$$$$\therefore \mathrm{\Theta }=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_{\mathrm{n}-1}-\gamma }{{\mathrm{n}}^2}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{({\mathrm{H}}_{\mathrm{n}}-\frac{1}{\mathrm{n}})}{{\mathrm{n}}^2}-\gamma \underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^3}-\gamma \underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2}$$$$=2\zeta \left(3\right)-\zeta \left(3\right)-\gamma \zeta \left(2\right)$$$$=\zeta \left(3\right)-\frac{\gamma {\pi }^2}{6}$$

Commented by mnjuly1970 last updated on 01/Jul/22

$$\mathrm{thanks}\mathrm{alot}$$

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