find the value of Σ_(n=2) ^∞ ((3n^2 +1)/((n−1)^3 (n+1)^3 ))


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Question Number 57489 by Abdo msup. last updated on 05/Apr/19

$f i n d t h e v a l u e o f \sum_{n = 2}^{\infty} \frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}}$
Commented by maxmathsup by imad last updated on 06/Apr/19
$we have (n+1)^3 −(n−1)^3 =n^3 +3n^2 +3n+1 −(n^3 −3n^2 +3n−1) =6n^2 +2 =2(3n^2 +1) ⇒3n^2 +1 =(1/2){(n+1)^3 −(n−1)^3 } ⇒ S =(1/2)Σ_(n=2) ^∞ (((n+1)^3 −(n−1)^3 )/((n−1)^3 (n+1)^3 )) =(1/2) Σ_(n=2) ^∞ (1/((n−1)^3 )) −(1/2) Σ_(n=2) ^∞ (1/((n+1)^3 )) but Σ_(n=2) ^∞ (1/((n−1)^3 )) =Σ_(n=1) ^∞ (1/n^3 ) =ξ(3) Σ_(n=2) ^∞ (1/((n+1)^3 )) =Σ_(n=3) ^∞ (1/n^3 ) =Σ_(n=1) ^∞ (1/n^3 ) −1−(1/8) =ξ(3)−(9/8) ⇒ S =(1/2)ξ(3)−(1/2)(ξ(3)−(9/8)) =(9/(16)) ★ S =(9/(16)) ★$
$w e h a v e {(n + 1)}^{3} - {(n - 1)}^{3} = n^{3} + 3 n^{2} + 3 n + 1 - (n^{3} - 3 n^{2} + 3 n - 1)$ $= 6 n^{2} + 2 = 2 (3 n^{2} + 1) \Rightarrow 3 n^{2} + 1 = \frac{1}{2} {{(n + 1)}^{3} - {(n - 1)}^{3}} \Rightarrow$ $S = \frac{1}{2} \sum_{n = 2}^{\infty} \frac{{(n + 1)}^{3} - {(n - 1)}^{3}}{{(n - 1)}^{3} {(n + 1)}^{3}} = \frac{1}{2} \sum_{n = 2}^{\infty} \frac{1}{{(n - 1)}^{3}} - \frac{1}{2} \sum_{n = 2}^{\infty} \frac{1}{{(n + 1)}^{3}}$ $b u t \sum_{n = 2}^{\infty} \frac{1}{{(n - 1)}^{3}} = \sum_{n = 1}^{\infty} \frac{1}{n^{3}} = ξ (3)$ $\sum_{n = 2}^{\infty} \frac{1}{{(n + 1)}^{3}} = \sum_{n = 3}^{\infty} \frac{1}{n^{3}} = \sum_{n = 1}^{\infty} \frac{1}{n^{3}} - 1 - \frac{1}{8} = ξ (3) - \frac{9}{8} \Rightarrow$ $S = \frac{1}{2} ξ (3) - \frac{1}{2} (ξ (3) - \frac{9}{8}) = \frac{9}{16}$ $★ S = \frac{9}{16} ★$
	Answered by Smail last updated on 06/Apr/19

	$F (n) \frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}} = \frac{a}{n - 1} + \frac{b}{{(n - 1)}^{2}} + \frac{c}{{(n - 1)}^{3}} + \frac{d}{n + 1} + \frac{e}{{(n + 1)}^{2}} + \frac{f}{{(n + 1)}^{3}}$ $F (- n) = - \frac{a}{n + 1} + \frac{b}{{(n + 1)}^{2}} - \frac{c}{{(n + 1)}^{3}} - \frac{d}{(n - 1)} + \frac{e}{{(n - 1)}^{2}} - \frac{f}{{(n - 1)}^{3}}$ $F (- n) = F (n)$ $b = e; a = - d; c = - f = \frac{1}{2}$ $F (0) = - 1 = - a + b - c + d + e + f$ $b - a - c - a + b - c = - 1 \Leftrightarrow b - a - c = \frac{- 1}{2} \Leftrightarrow b = a$ $F (2) = \frac{13}{27} = a + a + c - \frac{a}{3} + \frac{a}{9} - \frac{c}{27}$ $= a (2 - \frac{2}{9}) + \frac{1}{2} (1 - \frac{1}{27}) = \frac{13}{27}$ $\frac{16}{9} a + \frac{26}{2 \times 27} = \frac{13}{27} \Leftrightarrow a = 0$ $a = b = e = d = 0$ $\frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}} = \frac{1}{2} (\frac{1}{{(n - 1)}^{3}} - \frac{1}{{(n + 1)}^{3}})$ $\underset{n = 2}{\sum^{\infty}} \frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}} = \frac{1}{2} \underset{n = 2}{\sum^{\infty}} (\frac{1}{{(n - 1)}^{3}} - \frac{1}{{(n + 1)}^{3}})$ $l e t m + 1 = n - 1$ $\underset{n = 2}{\sum^{\infty}} \frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}} = \frac{1}{2} \underset{m = 0}{\sum^{\infty}} \frac{1}{{(m + 1)}^{3}} - \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}}$ $= \frac{1}{2} (\underset{n = 2}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}} + 1 + \frac{1}{2^{3}}) - \frac{1}{2} \underset{n = 2}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}}$ $= \frac{1}{2} (\frac{2^{3} + 1}{2^{3}})$ $\underset{n = 2}{\sum^{\infty}} \frac{3 n^{2} + 1}{{(n - 1)}^{3} {(n + 1)}^{3}} = \frac{9}{16}$
	Commented by maxmathsup by imad last updated on 06/Apr/19

	$s i r S m a i l y o u a n s w e r i s c o r r e c t t h a n k s a l o t .$
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