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find-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
find Σ_(n=1) ^(+∞) ((4n)/((2n−1)^2 (2n+1)^2 ))
find∑n=1+∞4n(2n−1)2(2n+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)
weseethat2n+1+2n−1=4nsothefractionF(x)=4x(2x−1)2(2x+1)2havedecompositionatformF(x)=a(2x−1)2+b(2x+1)2a=limx→12(2x−1)2F(x)=24=12b=limx→−12(2x+1)2F(x)=−12Sn=∑k=1n4k(2k−1)2(2k+1)2=12∑k=1n1(2k−1)2−12∑k=1n1(2k+1)2=−12∑k=1n(uk+1−uk)(uk=1(2k−1)2)=−12{un+1−u1}=−12{1(2n+1)2−1}=12−12(2n+1)2⇒limn→+∞Sn=12
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
=12{∑+∞n=11(2n−1)2−1(2n+1)2}12{(112+132+152+….)−(132+152+172+…)}=12imadea?mistake…(2n+1)2−(2n−1)2=4×2n×1=8nbutinproblem4n(2n+1)2(2n−1)2so12{1(2n−1)2−1(2n+1)2}so12factorshouldbethere…ormethld2s=112−132+132−152….….+1(2n−1)2−1(2n+1)22Sn=1−1(2n+1)2Whenn→∞Sn=1=12ANS

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