find the value of Σ_(n=1) ^∞ (n/((4n^2 −1)^2 )) .


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Question Number 45963 by maxmathsup by imad last updated on 19/Oct/18

$f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{n}{{(4 n^{2} - 1)}^{2}} .$
Commented by maxmathsup by imad last updated on 23/Oct/18
$let decompose F(n)=(n/((4n^2 −1)^2 )) ⇒F(x)=(n/((2n−1)^2 (2n+1)^2 )) =(1/8){ (1/((2n−1)^2 )) −(1/((2n+1)^2 ))} ⇒Σ_(n=1) ^∞ (n/((4n^2 −1)^2 )) =(1/8)Σ_(n=1) ^∞ (1/((2n−1)^2 )) −(1/8) Σ_(n=1) ^∞ (1/((2n+1)^2 )) but Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (1/((2n+1)^2 )) =(π^2 /8) −1 and Σ_(n=1) ^∞ (1/((2n−1)^2 )) =_(n=k+1) Σ_(k=0) ^∞ (1/((2k+1)^2 )) =(π^2 /8) ⇒ S =(1/8){(π^2 /8) +(π^2 /8) −1} =(1/8){(π^2 /4)−1}=(π^2 /(32)) −(1/8) .$
$l e t d e c o m p o s e F (n) = \frac{n}{{(4 n^{2} - 1)}^{2}} \Rightarrow F (x) = \frac{n}{{(2 n - 1)}^{2} {(2 n + 1)}^{2}}$ $= \frac{1}{8} {\frac{1}{{(2 n - 1)}^{2}} - \frac{1}{{(2 n + 1)}^{2}}} \Rightarrow \sum_{n = 1}^{\infty} \frac{n}{{(4 n^{2} - 1)}^{2}}$ $= \frac{1}{8} \sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}} - \frac{1}{8} \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{2}} b u t$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}$ $\Rightarrow \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8} - 1 a n d \sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}} =_{n = k + 1} \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{2}}$ $= \frac{π^{2}}{8} \Rightarrow S = \frac{1}{8} {\frac{π^{2}}{8} + \frac{π^{2}}{8} - 1} = \frac{1}{8} {\frac{π^{2}}{4} - 1} = \frac{π^{2}}{32} - \frac{1}{8} .$
	Answered by Smail last updated on 22/Oct/18

	$A = \underset{n = 1}{\sum^{\infty}} \frac{n}{{(4 n^{2} - 1)}^{2}} = \underset{n = 1}{\sum^{\infty}} \frac{n}{{(2 n - 1)}^{2} {(2 n + 1)}^{2}}$ $\frac{n}{{(2 n - 1)}^{2} {(2 n + 1)}^{2}} = \frac{1}{8} (\frac{1}{{(2 n - 1)}^{2}} - \frac{1}{{(2 n + 1)}^{2}}$ $A = \frac{1}{8} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n - 1)}^{2}} - \frac{1}{8} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}}$ $= \frac{1}{8} [\underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n - 1)}^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{2}} - (\underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n + 1)}^{2}} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(2 n)}^{2}})]$ $= \frac{1}{8} [\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} - \underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2}}]$ $= \frac{1}{8} (1) = \frac{1}{8}$
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