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


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Question Number 36819 by maxmathsup by imad last updated on 06/Jun/18

$f i n d t h e v a l u e o f t h e s u m \sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2} {(2 n + 1)}^{2}}$
Commented by maxmathsup by imad last updated on 03/Aug/18
$let S_n = Σ_(k=1) ^n (1/((2k−1)^2 (2k+1)^2 )) ⇒S=lim_(n→+∞) S_n let decompose F(x)=(1/((2x−1)^2 (2x+1)^2 )) F(x)= (a/(2x−1)) +(b/((2x−1)^2 )) +(c/(2x+1)) +(d/((2x+1)^2 )) b =lim_(x→(1/2)) (2x−1)^2 F(x) = (1/4) d=lim_(x→−(1/2)) (2x+1)^2 F(x)=(1/4) ⇒F(x)=(a/(2x−1)) +(1/(4(2x−1)^2 )) +(c/((2x+1))) +(1/(4(2x+1)^2 )) lim_(x→+∞) xF(x)=0 =(a/2) +(c/2) ⇒c=−a ⇒ F(x)=(a/(2x−1)) −(a/(2x+1)) + (1/(4(2x−1)^2 )) +(1/(4(2x+1)^2 )) F(0) =1 =−2a +(1/2) ⇒2a =−(1/2) ⇒a =−(1/4) ⇒ F(x)=(1/(4(2x+1))) −(1/(4(2x−1))) +(1/(4(2x−1)^2 )) +(1/(4(2x+1)^2 )) S_n =Σ_(k=1) ^n F(k) =(1/4){ Σ_(k=1) ^n (1/(2k+1)) −Σ_(k=1) ^n (1/(2k−1)) +Σ_(k=1) ^n (1/((2k−1)^2 )) +Σ_(k=1) ^n (1/((2k+1)^2 ))}but Σ_(k=1) ^n ((1/(2k+1)) −(1/(2k−1))) =Σ_(k=1) ^n (u_n −u_(n−1) ) (u_n =(1/(2n+1))) =u_n −u_0 =(1/(2n+1)) −1 also Σ_(k=1) ^n (1/((2k−1)^2 )) =_(k=j+1) Σ_(j=0) ^(n−1) (1/((2j+1)^2 )) →Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(k=1) ^n (1/((2k+1)^2 )) =Σ_(k=0) ^n (1/((2k+1)^2 )) −1 →Σ_(n=0) ^∞ (1/((2n+1)^2 )) −1 we have (π^2 /6) =Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4) (π^2 /6) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /6) −(π^2 /(24)) =((3π^2 )/(24)) =(π^2 /8) ⇒ lim_ S_n =(1/4){−1 +(π^2 /8) +(π^2 /8) −1} =(1/4){(π^2 /4) −2} =(π^2 /(16)) −(1/2) S=(π^2 /(16)) −(1/2) .$
$l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2} {(2 k + 1)}^{2}} \Rightarrow S = {l i m}_{n \to + \infty} S_{n} l e t d e c o m p o s e$ $F (x) = \frac{1}{{(2 x - 1)}^{2} {(2 x + 1)}^{2}}$ $F (x) = \frac{a}{2 x - 1} + \frac{b}{{(2 x - 1)}^{2}} + \frac{c}{2 x + 1} + \frac{d}{{(2 x + 1)}^{2}}$ $b = {l i m}_{x \to \frac{1}{2}} {(2 x - 1)}^{2} F (x) = \frac{1}{4}$ $d = {l i m}_{x \to - \frac{1}{2}} {(2 x + 1)}^{2} F (x) = \frac{1}{4} \Rightarrow F (x) = \frac{a}{2 x - 1} + \frac{1}{4 {(2 x - 1)}^{2}} + \frac{c}{(2 x + 1)} + \frac{1}{4 {(2 x + 1)}^{2}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{2} + \frac{c}{2} \Rightarrow c = - a \Rightarrow$ $F (x) = \frac{a}{2 x - 1} - \frac{a}{2 x + 1} + \frac{1}{4 {(2 x - 1)}^{2}} + \frac{1}{4 {(2 x + 1)}^{2}}$ $F (0) = 1 = - 2 a + \frac{1}{2} \Rightarrow 2 a = - \frac{1}{2} \Rightarrow a = - \frac{1}{4} \Rightarrow$ $F (x) = \frac{1}{4 (2 x + 1)} - \frac{1}{4 (2 x - 1)} + \frac{1}{4 {(2 x - 1)}^{2}} + \frac{1}{4 {(2 x + 1)}^{2}}$ $S_{n} = \sum_{k = 1}^{n} F (k) = \frac{1}{4} {\sum_{k = 1}^{n} \frac{1}{2 k + 1} - \sum_{k = 1}^{n} \frac{1}{2 k - 1} + \sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} + \sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}}} b u t$ $\sum_{k = 1}^{n} (\frac{1}{2 k + 1} - \frac{1}{2 k - 1}) = \sum_{k = 1}^{n} (u_{n} - u_{n - 1}) (u_{n} = \frac{1}{2 n + 1})$ $= u_{n} - u_{0} = \frac{1}{2 n + 1} - 1 a l s o$ $\sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} =_{k = j + 1} \sum_{j = 0}^{n - 1} \frac{1}{{(2 j + 1)}^{2}} \to \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ $\sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}} = \sum_{k = 0}^{n} \frac{1}{{(2 k + 1)}^{2}} - 1 \to \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} - 1$ $w e h a v e \frac{π^{2}}{6} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{4 n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{1}{4} \frac{π^{2}}{6} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}$ $\Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{π^{2}}{6} - \frac{π^{2}}{24} = \frac{3 π^{2}}{24} = \frac{π^{2}}{8} \Rightarrow$ ${l i m}_{} S_{n} = \frac{1}{4} {- 1 + \frac{π^{2}}{8} + \frac{π^{2}}{8} - 1} = \frac{1}{4} {\frac{π^{2}}{4} - 2} = \frac{π^{2}}{16} - \frac{1}{2}$ $S = \frac{π^{2}}{16} - \frac{1}{2} .$
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