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Question Number 158531 by amin96 last updated on 05/Nov/21

$$\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathit{n}}^2{(2\mathit{n}+1)}^3}=?$$

Answered by qaz last updated on 06/Nov/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{n}}^2{(2\mathrm{n}+1)}^3}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{{8\mathrm{n}}^2}-\frac{3}{4\mathrm{n}}+\frac{1}{2{(2\mathrm{n}+1)}^3}+\frac{1}{{(2\mathrm{n}+1)}^2}+\frac{3}{2(2\mathrm{n}+1)})$$$$=\frac{1}{8}\zeta \left(2\right)+\frac{7}{16}\zeta \left(3\right)+\frac{3}{4}\zeta \left(2\right)+\frac{3}{4}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}+\frac{1}{2}}-\frac{1}{\mathrm{n}})$$$$=\frac{7}{8}\zeta \left(2\right)+\frac{7}{16}\zeta \left(3\right)-\frac{3}{4}{\mathrm{H}}_{1/2}$$$$=\frac{7}{8}\zeta \left(2\right)+\frac{7}{16}\zeta \left(3\right)-\frac{3}{2}+\frac{3}{2}\ln 2$$

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