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Question Number 65673 by mathmax by abdo last updated on 01/Aug/19

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

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

Commented by mathmax by abdo last updated on 02/Aug/19

let S =Σ_(n=2) ^∞ (((−1)^n )/((n^2 −1)^2 )) let decompose F(x) =(1/((x^2 −1)^2 )) F(x) =(1/((x−1)^2 (x+1)^2 )) =(a/(x−1))+(b/((x−1)^2 )) +(c/((x+1))) +(d/((x+1)^2 )) b =lim_(x→1) (x−1)^2 F(x) =(1/4) d =lim_(x→−1) (x+1)^2 F(x) =(1/4) ⇒ F(x) =(a/(x−1)) +(1/(4(x−1)^2 )) +(c/(x+1)) +(1/(4(x+1)^2 )) lim_(x→+∞) xF(x) =0 =a+c ⇒c=−a ⇒ F(x) =(a/(x−1)) +(1/(4(x−1)^2 ))−(a/(x+1)) +(1/(4(x+1)^2 )) F(0) =1 =−a +(1/4) −a +(1/4) =−2a+(1/2) ⇒2a =(1/2)−1 =−(1/2) ⇒ a =−(1/4) ⇒F(x) =((−1)/(4(x−1))) +(1/(4(x−1)^2 )) +(1/(4(x+1))) +(1/(4(x+1)^2 )) S =Σ_(n=2) ^∞ (−1)^n F(n) =−(1/4)Σ_(n=2) ^∞ (((−1)^n )/(n−1)) +(1/4)Σ_(n=2) ^∞ (((−1)^n )/((n−1)^2 )) +(1/4)Σ_(n=2) ^∞ (((−1)^n )/((n+1))) +(1/4)Σ_(n=2) ^∞ (((−1)^n )/((n+1)^2 )) Σ_(n=2) ^∞ (((−1)^n )/(n−1)) =Σ_(n=1) ^∞ (((−1)^(n+1) )/n) =−Σ_(n=1) ^∞ (((−1)^n )/n) =ln(2) Σ_(n=2) ^∞ (((−1)^n )/(n+1)) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −1+(1/2) =ln(2)−(1/2) Σ_(n=2) ^∞ (((−1)^n )/((n−1)^2 )) =Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) but we have Σ_(n=1) ^∞ (((−1)^n )/n^x ) =(2^(1−x) −1)ξ(x) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(1−2) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(n=2) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=3) ^∞ (((−1)^n )/n^2 ) =−{Σ_(n=1) ^∞ (((−1)^n )/n^2 )+1−(1/4)} =(3/4)−(−(π^2 /(12)))=(3/4) +(π^2 /(12)) ⇒ S =−(1/4)ln(2) +(1/4)((π^2 /(12))) +(1/4)(ln2−(1/2)) +(1/4)((3/4)+(π^2 /(12))) =(π^2 /(48))−(1/8) +(3/(16)) +(π^2 /(48)) =(π^2 /(24)) +(1/(16))

$${let}\:{S}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:{let}\:{decompose}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{{a}}{{x}−\mathrm{1}}+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({x}+\mathrm{1}\right)}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{{a}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=−{a}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−{a}\:+\frac{\mathrm{1}}{\mathrm{4}}\:=−\mathrm{2}{a}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{a}\:=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${a}\:=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{F}\left({x}\right)\:=\frac{−\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {F}\left({n}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:={ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\:\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${but}\:{we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}^{\mathrm{2}} } \\ $$$$=−\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right\}\:=\frac{\mathrm{3}}{\mathrm{4}}−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow \\ $$$${S}\:=−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left({ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{8}}\:+\frac{\mathrm{3}}{\mathrm{16}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:+\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$ \\ $$

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