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


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Question Number 38323 by math khazana by abdo last updated on 24/Jun/18

$f i n d \sum_{n = 1}^{+ \infty} \frac{4 n}{{(2 n - 1)}^{2} {(2 n + 1)}^{2}}$
Commented by math khazana by abdo last updated on 25/Jun/18
$we see that 2n+1+2n−1=4n so the fraction F(x)=((4x)/((2x−1)^2 (2x+1)^2 )) have decomposition at form F(x)= (a/((2x−1)^2 )) + (b/((2x+1)^2 )) a=lim_(x→(1/2)) (2x−1)^2 F(x)= (2/4) =(1/2) b=lim_(x→−(1/2)) (2x+1)^2 F(x)=−(1/2) S_n =Σ_(k=1) ^n ((4k)/((2k−1)^2 (2k+1)^2 )) =(1/2)Σ_(k=1) ^n (1/((2k−1)^2 )) −(1/2)Σ_(k=1) ^n (1/((2k+1)^2 )) =−(1/2)Σ_(k=1) ^n (u_(k+1) −u_k ) (u_k = (1/((2k−1)^2 ))) =−(1/2){u_(n+1) −u_1 }=−(1/2){ (1/((2n+1)^2 )) −1} = (1/2) −(1/(2(2n+1)^2 )) ⇒lim_(n→+∞) S_n =(1/2)$
$w e s e e t h a t 2 n + 1 + 2 n - 1 = 4 n s o t h e f r a c t i o n$ $F (x) = \frac{4 x}{{(2 x - 1)}^{2} {(2 x + 1)}^{2}} h a v e d e c o m p o s i t i o n$ $a t f o r m F (x) = \frac{a}{{(2 x - 1)}^{2}} + \frac{b}{{(2 x + 1)}^{2}}$ $a = {l i m}_{x \to \frac{1}{2}} {(2 x - 1)}^{2} F (x) = \frac{2}{4} = \frac{1}{2}$ $b = {l i m}_{x \to - \frac{1}{2}} {(2 x + 1)}^{2} F (x) = - \frac{1}{2}$ $S_{n} = \sum_{k = 1}^{n} \frac{4 k}{{(2 k - 1)}^{2} {(2 k + 1)}^{2}}$ $= \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{{(2 k - 1)}^{2}} - \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{{(2 k + 1)}^{2}}$ $= - \frac{1}{2} \sum_{k = 1}^{n} (u_{k + 1} - u_{k}) (u_{k} = \frac{1}{{(2 k - 1)}^{2}})$ $= - \frac{1}{2} {u_{n + 1} - u_{1}} = - \frac{1}{2} {\frac{1}{{(2 n + 1)}^{2}} - 1}$ $= \frac{1}{2} - \frac{1}{2 {(2 n + 1)}^{2}} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18
	$=(1/2){Σ_(n=1) ^(+∞) (1/((2n−1)^2 ))−(1/((2n+1)^2 ))} (1/2){((1/1^2 )+(1/3^2 )+(1/5^2 )+....)−((1/3^2 )+(1/5^2 )+(1/7^2 )+...)} =(1/2) i made a?mistake... (2n+1)^2 −(2n−1)^2 =4×2n×1=8n but in problem ((4n)/((2n+1)^2 (2n−1)^2 )) so(1/2){(1/((2n−1)^2 ))−(1/((2n+1)^2 ))} so (1/2) factor should be there... or methld 2s=(1/1^2 )−(1/3^2 ) +(1/3^2 )−(1/5^2 ) .... .... +(1/((2n−1)^2 ))−(1/((2n+1)^2 )) 2S_n =1−(1/((2n+1)^2 )) When n→∞S_n =1 =(1/2)ANS$
	$= \frac{1}{2} {\underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(2 n - 1)}^{2}} - \frac{1}{{(2 n + 1)}^{2}}}$ $\frac{1}{2} {(\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . .) - (\frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + . . .)}$ $= \frac{1}{2} i m a d e a ? m i s t a k e . . .$ ${(2 n + 1)}^{2} - {(2 n - 1)}^{2} = 4 \times 2 n \times 1 = 8 n$ $b u t i n p r o b l e m \frac{4 n}{{(2 n + 1)}^{2} {(2 n - 1)}^{2}}$ $s o \frac{1}{2} {\frac{1}{{(2 n - 1)}^{2}} - \frac{1}{{(2 n + 1)}^{2}}} s o \frac{1}{2} f a c t o r s h o u l d$ $b e t h e r e . . .$ $o r m e t h l d$ $2 s = \frac{1}{1^{2}} - \frac{1}{3^{2}}$ $+ \frac{1}{3^{2}} - \frac{1}{5^{2}}$ $. . . .$ $. . . .$ $+ \frac{1}{{(2 n - 1)}^{2}} - \frac{1}{{(2 n + 1)}^{2}}$ $2 S_{n} = 1 - \frac{1}{{(2 n + 1)}^{2}} W h e n n \to \infty S_{n} = 1$ $= \frac{1}{2} A N S$
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