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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 ))

$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$

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

$${let}\:{named}\:{it}\:{S} \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left[\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\right]^{\mathrm{2}} }\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\left[\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}\right]^{\mathrm{2}} \\ $$$$\mathrm{2}{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left[\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right]^{\mathrm{2}} \\ $$$$\:\mathrm{2}{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left[\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right] \\ $$$$\mathrm{2}{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{2}{S}=−\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} }\:−{A}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{with}\:\:{A}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}\: \\ $$$$\mathrm{2}{S}=−\mathrm{1}−{A}\: \\ $$$${let}\:{state}\:\:{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)}\:\:\:{we}\:{have}\:\:{A}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{U}_{{n}_{} } \\ $$$${U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${So}\:\:{A}=\mathrm{1}\:\:{then}\:\:{S}=−\mathrm{2} \\ $$

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