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Question Number 74351 by mathmax by abdo last updated on 22/Nov/19
find the value of  Σ_(n=1) ^(+∞)    (((−1)^n )/((4n^2 −1)^2 ))
findthevalueofn=1+(1)n(4n21)2
Commented by ~blr237~ last updated on 23/Nov/19
let named it S  S=Σ_(n=1) ^∞  (((−1)^n )/([(2n+1)(2n−1)]^2 )) =Σ_(n=1) ^∞ (((−1)^n )/2)[(((2n+1)−(2n−1))/((2n+1)(2n−1)))]^2   2S=Σ_(n=1) ^∞ (−1)^n [(1/(2n−1))−(1/(2n+1))]^2    2S=Σ_(n=1) ^∞ (−1)^n [(1/((2n−1)^2 ))−(2/((2n+1)(2n−1)))+(1/((2n+1)^2 ))]  2S=Σ_(n=1) ^∞ (((−1)^n )/((2n−1)^2 )) −Σ_(n=1) ^∞ (2/((2n+1)(2n−1)))+Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 ))   2S=−Σ_(m=0) ^∞ (((−1)^m )/((2m+1)^2 )) −A+Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 ))   with  A=Σ_(n=1) ^∞ (2/((2n+1)(2n−1)))   2S=−1−A   let state  U_n =Σ_(k=1) ^n (2/((2k+1)(2k−1)))   we have  A=lim_(n→∞)  U_n_    U_n =Σ_(k=1) ^n ((1/(2k−1))−(1/(2k+1)))=Σ_(k=0) ^(n−1) (1/(2k+1)) −Σ_(k=1) ^n (1/(2k+1))=1−(1/(2n+1))  So  A=1  then  S=−2
letnameditSS=n=1(1)n[(2n+1)(2n1)]2=n=1(1)n2[(2n+1)(2n1)(2n+1)(2n1)]22S=n=1(1)n[12n112n+1]22S=n=1(1)n[1(2n1)22(2n+1)(2n1)+1(2n+1)2]2S=n=1(1)n(2n1)2n=12(2n+1)(2n1)+n=1(1)n(2n+1)22S=m=0(1)m(2m+1)2A+n=1(1)n(2n+1)2withA=n=12(2n+1)(2n1)2S=1AletstateUn=nk=12(2k+1)(2k1)wehaveA=limnUnUn=nk=1(12k112k+1)=n1k=012k+1nk=112k+1=112n+1SoA=1thenS=2

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