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Question Number 143031 by rs4089 last updated on 09/Jun/21

Answered by mnjuly1970 last updated on 09/Jun/21

$$\frac{{\pi }^2}{6}+{ln}^2\left(2\right)....$$

Answered by Dwaipayan Shikari last updated on 09/Jun/21

$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_n^2{x}^n=H_1^{2}x+H_2^2{x}^2+H_3{x}^3+...$$$$S(1-x)=x+(H_2^2-H_1^2)x^2+(H_3^2-H_2^2)x^3+...$$$$S(1-x)=x+\frac{H_2+H_1}{2}{x}^2+\frac{H_2+H_3}{3}{x}^3+...H_{n-1}=H_n-\frac{1}{n}$$$$S(1-x)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n+H_{n-1}}{n}{x}^n=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2H_n-\frac{1}{n}}{n}{x}^n$$$$\Rightarrow S(1-x)=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n}{x}^n-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_n{x}^{n-1}=-\frac{log(1-x)}{x(1-x)}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n{2}^n}=-{\int }_0^{1/2}\frac{log(1-x)}{x}dx-{\int }_0^{1/2}\frac{log(1-x)}{1-x}dx$$$$={Li}_2\left(\frac{1}{2}\right)+\frac{1}{2}{log}^2\left(2\right)$$$${Li}_2(1-x)+{Li}_2\left(x\right)=\frac{{\pi }^2}{6}-\frac{1}{2}log\left(x\right)log(1-x)$$$$\Rightarrow {Li}_2\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}-\frac{1}{2}{log}^2\left(2\right)$$$$S(1-\frac{1}{2})=2({Li}_2\left(\frac{1}{2}\right)+{log}^2\left(2\right))-{Li}_2\left(\frac{1}{2}\right)$$$$S=2{Li}_2\left(\frac{1}{2}\right)+2{log}^2\left(2\right)=\frac{{\pi }^2}{6}+{log}^2\left(2\right)$$$$$$

Commented by mnjuly1970 last updated on 09/Jun/21

$$greate....$$

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