calculate Σ_(n=1) ^∞ (1/((4n^2 −1)^2 ))


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Question Number 45594 by maxmathsup by imad last updated on 14/Oct/18

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{1}{{(4 n^{2} - 1)}^{2}}$
Commented by maxmathsup by imad last updated on 18/Oct/18
$let S_n =Σ_(k=1) ^n (1/((4k^2 −1)^2 )) ⇒S_n =Σ_(k=1) ^n (1/((2k−1)^2 (2k+1)^2 )) let decompose F(x)=(1/((2x−1)^2 (2x+1)^2 )) ⇒F(x)=(a/(2x−1)) +(b/((2x−1)^2 )) +(c/(2x+1)) +(d/((2x+1)^2 )) b =lim_(x→(1/2)) (2x−1)^2 F(x)=(1/4) d =lim_(x→−(1/2)) (2x+1)^2 F(x) =(1/4) ⇒F(x)=(a/(2x−1)) +(1/(4(2x−1)^2 )) +(c/((2x+1))) +(d/(4(2x+)^2 )) lim_(x→+∞) x F(x)=0 =(a/2) +(c/2) ⇒c=−a ⇒ F(x)=(a/(2x−1)) −(a/(2x+1)) + (1/(4(2x−1)^2 )) +(1/(4(2x+1)^2 )) F(0)=1 =−2a +(1/2) ⇒(1/2) =−2a ⇒a=−(1/4) ⇒ F(x)=(1/(4(2x+1)))−(1/(4(2x−1))) +(1/(4(2x−1)^2 )) +(1/(4(2x+1)^2 )) ⇒ S_n =Σ_(k=1) ^n F(k)=(1/4) Σ_(k=1) ^n (1/(2k+1)) −(1/4)Σ_(k=1) ^n (1/(2k−1)) +(1/4)Σ_(k=1) ^n (1/((2k−1)^2 )) +(1/4)Σ_(k=1) ^n (1/((2k+1)^2 )) (1/4)Σ_(k=1) ^n {(1/(2k+1))−(1/(2k−1))} =(1/4)Σ_(k=1) ^n v_k −v_(k−1) =(1/4)(v_n −v_0 ) (v_n =(1/(2k+1))) =(1/4)((1/(2n+1)) −1) →−(1/4)(n→+∞) we have(1/4) {Σ_(k=1) ^n (1/((2k−1)^2 )) +Σ_(k=1) ^n (1/((2k+1)^2 ))}→(1/4){Σ_(n=1) ^∞ (1/((2n−1)^2 )) +Σ_(n=1) ^∞ (1/((2n+1)^2 ))} but Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (1/((2n+1)^2 )) =(π^2 /8) −1 Σ_(n=1) ^∞ (1/((2n−1)^2 )) =_(n=k+1) Σ_(k=0) ^∞ (1/((2k+1)^2 )) =(π^2 /8) ⇒ lim_(n→+∞) S_n =−(1/4) +(1/4){(π^2 /8)−1 +(π^2 /8)}=(1/4){(π^2 /4) −1)−(1/4) =(π^2 /(16)) −(1/2) .$
$l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{{(4 k^{2} - 1)}^{2}} \Rightarrow S_{n} = \sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2} {(2 k + 1)}^{2}} l e t d e c o m p o s e$ $F (x) = \frac{1}{{(2 x - 1)}^{2} {(2 x + 1)}^{2}} \Rightarrow F (x) = \frac{a}{2 x - 1} + \frac{b}{{(2 x - 1)}^{2}} + \frac{c}{2 x + 1} + \frac{d}{{(2 x + 1)}^{2}}$ $b = {l i m}_{x \to \frac{1}{2}} {(2 x - 1)}^{2} F (x) = \frac{1}{4}$ $d = {l i m}_{x \to - \frac{1}{2}} {(2 x + 1)}^{2} F (x) = \frac{1}{4} \Rightarrow F (x) = \frac{a}{2 x - 1} + \frac{1}{4 {(2 x - 1)}^{2}} + \frac{c}{(2 x + 1)} + \frac{d}{4 {(2 x +)}^{2}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{2} + \frac{c}{2} \Rightarrow c = - a \Rightarrow$ $F (x) = \frac{a}{2 x - 1} - \frac{a}{2 x + 1} + \frac{1}{4 {(2 x - 1)}^{2}} + \frac{1}{4 {(2 x + 1)}^{2}}$ $F (0) = 1 = - 2 a + \frac{1}{2} \Rightarrow \frac{1}{2} = - 2 a \Rightarrow a = - \frac{1}{4} \Rightarrow$ $F (x) = \frac{1}{4 (2 x + 1)} - \frac{1}{4 (2 x - 1)} + \frac{1}{4 {(2 x - 1)}^{2}} + \frac{1}{4 {(2 x + 1)}^{2}} \Rightarrow$ $S_{n} = \sum_{k = 1}^{n} F (k) = \frac{1}{4} \sum_{k = 1}^{n} \frac{1}{2 k + 1} - \frac{1}{4} \sum_{k = 1}^{n} \frac{1}{2 k - 1} + \frac{1}{4} \sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} + \frac{1}{4} \sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}}$ $\frac{1}{4} \sum_{k = 1}^{n} {\frac{1}{2 k + 1} - \frac{1}{2 k - 1}} = \frac{1}{4} \sum_{k = 1}^{n} v_{k} - v_{k - 1} = \frac{1}{4} (v_{n} - v_{0}) (v_{n} = \frac{1}{2 k + 1})$ $= \frac{1}{4} (\frac{1}{2 n + 1} - 1) \to - \frac{1}{4} (n \to + \infty)$ $w e h a v e \frac{1}{4} {\sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} + \sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}}} \to \frac{1}{4} {\sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}} + \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{2}}}$ $b u t \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{8} - 1$ $\sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}} =_{n = k + 1} \sum_{k = 0}^{\infty} \frac{1}{{(2 k + 1)}^{2}} = \frac{π^{2}}{8} \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = - \frac{1}{4} + \frac{1}{4} {\frac{π^{2}}{8} - 1 + \frac{π^{2}}{8}} = \frac{1}{4} {\frac{π^{2}}{4} - 1) - \frac{1}{4}$ $= \frac{π^{2}}{16} - \frac{1}{2} .$
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