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Question Number 166254 by mnjuly1970 last updated on 16/Feb/22

$$$$$$\mathrm{\Theta }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{H_n}{n.(n+1)}\stackrel{?}{=}\frac{{\pi }^2}{6}$$$$----+$$

Answered by Kamel_Ben last updated on 16/Feb/22

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

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