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Question Number 159465 by mnjuly1970 last updated on 17/Nov/21

$$$$$$simplify$$$$$$$$\xi :=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{\underset{k=1}{\sum ^n}{k}^3}\right)=?$$$$$$

Answered by Ar Brandon last updated on 17/Nov/21

$$\xi =\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{\underset{k=1}{\sum ^n}{k}^3}\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{4}{n^2{(n+1)}^2}\right)$$$$\frac{1}{n^2{(n+1)}^2}=\frac{1}{n^2(n+1)}-\frac{1}{n{(n+1)}^2}$$$$=\frac{1}{n^2}-\frac{1}{n(n+1)}-\frac{1}{n(n+1)}+\frac{1}{{(n+1)}^2}$$$$=\frac{1}{n^2}-\frac{2}{n}+\frac{2}{n+1}+\frac{1}{{(n+1)}^2}$$$$\xi =4\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n^2}-\frac{2}{n}+\frac{2}{n+1}+\frac{1}{{(n+1)}^2})$$$$=4(\zeta \left(2\right)-2H_n+2H_n-2+\zeta \left(2\right)-1)=4(2\zeta \left(2\right)-3)$$$$=4(\frac{{\pi }^2}{3}-3)=\frac{4{\pi }^2}{3}-12=\frac{4}{3}(\pi -3)(\pi +3)$$

Commented by mnjuly1970 last updated on 17/Nov/21

$$thanksalotsirbrandon.mercey$$

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