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Question Number 151630 by mathlove last updated on 22/Aug/21

Answered by qaz last updated on 22/Aug/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}$$$$={\int }_0^1\frac{1}{1-\mathrm{x}}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1-{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2}\mathrm{dx}$$$$={\int }_0^1\frac{\frac{{\pi }^2}{6}-{\mathrm{Li}}_2\left(\mathrm{x}\right)}{1-\mathrm{x}}\mathrm{dx}$$$$={\int }_0^1\frac{\frac{{\pi }^2}{6}-{\mathrm{Li}}_2(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=[\frac{{\pi }^2}{6}-{\mathrm{Li}}_2(1-\mathrm{x})]\mathrm{lnx}{\mid }_0^1-{\int }_0^1\frac{{\ln }^2\mathrm{x}}{1-\mathrm{x}}\mathrm{dx}$$$$=-\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}{\ln }^2\mathrm{xdx}$$$$=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(\mathrm{n}+1)}^3}$$$$=2\zeta \left(3\right)$$

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$$\bullet \underset{k=0}{\sum ^{n-1}}{x}^k=\frac{1-x^n}{1-x}$$$$\Rightarrow {\int }_0^1\underset{k=0}{\sum ^{n-1}}{x}^{k}dx={\int }_0^1\frac{1-x^n}{1-x}dx$$$${\left[\underset{k=0}{\sum ^{n-1}}\frac{x^{k+1}}{k+1}\right]}_0^1={\int }_0^1\frac{1-x^n}{1-x}dx$$$$\underset{k=0}{\sum ^{n-1}}\frac{1}{k+1}={\int }_0^1\frac{1-x^n}{1-x}dx$$$$\underset{k=1}{\sum ^n}\frac{1}{k}=H_n={\int }_0^1\frac{1-x^n}{1-x}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}{\int }_0^1\frac{1-x^n}{1-x}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2}\left(1\right)$$$$$$$$\bullet \underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\underset{m=1}{\sum ^{\mathrm{\infty }}}\underset{n=1}{\sum ^m}\frac{1}{m^{2}n}$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=n}{\sum ^{\mathrm{\infty }}}\frac{1}{m^{2}n}$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n^3}+\underset{m=n+1}{\sum ^{\mathrm{\infty }}}\frac{1}{m^{2}n})$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=n+1}{\sum ^{\mathrm{\infty }}}\frac{1}{m^{2}n}$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(m+n)}^{2}n}$$$$\mathrm{By}\mathrm{symmetry}:$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{{(m+n)}^{2}n}+\frac{1}{{(m+n)}^{2}m})$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{m+n}{mn{(m+n)}^2}$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{mn(m+n)}$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{1}{m(m+n)}$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}\underset{m=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{m}-\frac{1}{m+n})$$$$\mathrm{But}\mathrm{the}\mathrm{telescopic}\mathrm{sum}\underset{m=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{m}-\frac{1}{m+n})\mathrm{is}{H}_n.$$$$\underset{m=1}{\sum ^{\mathrm{\infty }}}\frac{H_m}{m^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2}=\zeta \left(3\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2}=2\zeta \left(3\right)\left(2\right)$$$$$$$$\bullet \left(1\right)\mathrm{and}\left(2\right)\mathrm{give}:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}{\int }_0^1\frac{1-x^n}{1-x}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n^2}=2\zeta \left(3\right)$$$$$$

Commented by Tawa11 last updated on 22/Aug/21

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by Tawa11 last updated on 22/Aug/21

$$\mathrm{Wow},\mathrm{great}\mathrm{sir}.$$

Commented by Tawa11 last updated on 22/Aug/21

$$\mathrm{Help}\mathrm{me}\mathrm{check}\mathrm{Q}151641$$

Commented by Tawa11 last updated on 22/Aug/21

$$\mathrm{And}\mathrm{Q}151636$$

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