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Question Number 143131 by mohammad17 last updated on 10/Jun/21

$$findthepartialsumsof\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+1)}$$

Answered by qaz last updated on 10/Jun/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2(\mathrm{n}+1)}$$$$={\int }_0^1\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2}\mathrm{dx}$$$$={\int }_0^1{\mathrm{Li}}_2\left(\mathrm{x}\right)\mathrm{dx}$$$$={\mathrm{xLi}}_2\left(\mathrm{x}\right){\mid }_0^1+{\int }_0^1\frac{\mathrm{xln}(1-\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=\frac{{\pi }^2}{6}+{\int }_0^1\ln (1-\mathrm{x})\mathrm{dx}$$$$=\frac{{\pi }^2}{6}-1$$

Commented by mohammad17 last updated on 11/Jun/21

$$thankyousirbutwhytheintervalofthe$$$$integralfrom\left(0to1\right)$$

Answered by Olaf_Thorendsen last updated on 10/Jun/21

$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+1)}$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1})$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}-\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n+1})$$$$\mathrm{S}=\frac{{\pi }^2}{6}-1$$

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