Σ_(n=1) ^∞ (1/(n(4n−1)^2 ))= ?


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Question Number 214302 by universe last updated on 04/Dec/24

$\underset{n = 1}{\sum^{\infty}} \frac{1}{n {(4 n - 1)}^{2}} = ?$
	Answered by MrGaster last updated on 24/Dec/24

	${(4 n - 1)}^{2} = 16 n^{2} - 8 n + 1$ $∴ \underset{n = 1}{\sum^{\infty}} \frac{1}{n (16 n^{2} - 8 n + 1)}$ $⇛ \underset{n = 1}{\sum^{\infty}} \frac{1}{16 n^{3} - 8 n^{2} + n}$ $\underset{n = 1}{\sum^{\infty}} (\frac{1}{16 n^{2}} - \frac{1}{16 n^{3}} + \frac{1}{16 n} - \frac{1}{8 n^{2}} + \frac{1}{8 n^{3}} - \frac{1}{8 n} + \frac{1}{n})$ $⇛ \underset{n = 1}{\sum^{\infty}} (\frac{1}{16 n^{2}} - \frac{1}{16 n^{3}} - \frac{1}{8 n^{2}} + \frac{1}{8 n^{3}} + \frac{7}{16 n})$ $⇛ \underset{n = 1}{\sum^{\infty}} (- \frac{1}{16 n^{3}} + \frac{1}{8 n^{3}} - \frac{1}{16 n^{2}} - \frac{1}{8 n^{2}} + \frac{7}{16 n})$ $⇛ \underset{n = 1}{\sum^{\infty}} (\frac{1}{8 n^{3}} - \frac{1}{16 n^{3}} - \frac{3}{16 n^{2}} + \frac{7}{16 n})$ $⇛ \underset{n = 1}{\sum^{\infty}} (\frac{1}{16 n^{3}} - \frac{3}{16 n^{2}} + \frac{7}{16 n})$ $⇛ \frac{1}{16} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n^{3}} - \frac{3}{n^{2}} + \frac{7}{n})$ $⇛ \frac{1}{16} (ζ (3) - 3 ζ (2) + 7 ζ (1))$ $⇛ \frac{1}{16} (ζ (3) - 3 ζ (2))$ $⇛ \frac{1}{16} (\underset{simplification}{\underset{⏟}{ζ (3)}} - \underset{simplification}{\underset{⏟}{3 \cdot \frac{π^{2}}{6}}})$ $= \frac{1}{16} (ζ (3) - \frac{π^{2}}{2})$
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