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

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

Answered by qaz last updated on 06/Nov/21

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

= \underset{n = 1}{\sum^{\infty}} (\frac{1}{{8 n}^{2}} - \frac{3}{4 n} + \frac{1}{2 {(2 n + 1)}^{3}} + \frac{1}{{(2 n + 1)}^{2}} + \frac{3}{2 (2 n + 1)})

= \frac{1}{8} ζ (2) + \frac{7}{16} ζ (3) + \frac{3}{4} ζ (2) + \frac{3}{4} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n + \frac{1}{2}} - \frac{1}{n})

= \frac{7}{8} ζ (2) + \frac{7}{16} ζ (3) - \frac{3}{4} H_{1 / 2}

= \frac{7}{8} ζ (2) + \frac{7}{16} ζ (3) - \frac{3}{2} + \frac{3}{2} \ln 2

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