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Question Number 31979 by abdo imad last updated on 17/Mar/18
calculate Σ_(n=2) ^∞ (1/((n^2 −1)^2 )) .
$${calculate}\:\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$
Commented by abdo imad last updated on 22/Mar/18
let put S_n =Σ_(k=2) ^n (1/((k^2 −1)^2 )) S_n = Σ_(k=2) ^n (1/((k−1)^2 (k+1)^2 )) let decompose 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) we have F(−x)=F(x)⇒ ((−a)/(x+1)) −(c/(x−1)) +(b/((x+1)^2 )) +(d/((x−1)^2 )) =F(x) ⇒c=−a ⇒F(0) F(x)= (a/(x−1)) −(a/(x+1)) + (1/(4(x−1)^2 )) + (1/(4(x+1)^2 )) F(0) = 1 =−2a +(1/2) ⇒2a =−(1/2) ⇒a=−(1/4) and F(x) = ((−1)/(4(x−1))) + (1/(4(x+1))) + (1/(4(x−1)^2 )) + (1/(4(x+1)^2 )) 4S_n =4Σ_(k=2) ^(n ) F(k)=−Σ_(k=2) ^n (1/(k−1)) +Σ_(k=2) ^n (1/(k+1)) +Σ_(k=2) ^n (1/((k−1)^2 )) +Σ_(k=2) ^n (1/((k+1)^2 )) =−Σ_(k=1) ^(n−1) (1/k) +Σ_(k=3) ^(n+1) (1/k) + Σ_(k=1) ^(n−1) (1/k^2 ) +Σ_(k=3) ^(n+1) (1/k^2 ) =−H_(n−1) +H_(n+1) −(3/2) + ξ_(n−1) (2) +ξ_(n+1) (2) −(5/4) =H_(n+1) −H_(n−1) −((11)/4) +ξ_(n−1) (2) +ξ_(n+1) (2) but H_(n+1 ) −H_(n−1) _(n→∞) →0 ξ_(n−1) (2) →(π^2 /6) ξ_(n+1) (2)→ (π^2 /6) ⇒ 4S_n → (π^2 /3) −((11)/4) ⇒ lim_(n→∞) S_n = (π^2 /(12)) −((11)/(16)) . let remember that H_n =Σ_(k=1) ^n (1/k) and ξ_n (x) =Σ_(k=1) ^n (1/k^x ) .
$${let}\:{put}\:{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}^{\mathrm{2}} \:−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{let}\:{decompose} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({x}+\mathrm{1}\right)}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${d}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:{we}\:{have}\:{F}\left(−{x}\right)={F}\left({x}\right)\Rightarrow \\ $$$$\frac{−{a}}{{x}+\mathrm{1}}\:−\frac{{c}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{d}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:={F}\left({x}\right)\:\Rightarrow{c}=−{a}\:\Rightarrow{F}\left(\mathrm{0}\right) \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:−\frac{{a}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:=−\mathrm{2}{a}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{a}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{4}}\:{and} \\ $$$${F}\left({x}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}{S}_{{n}} =\mathrm{4}\sum_{{k}=\mathrm{2}} ^{{n}\:} \:{F}\left({k}\right)=−\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:+\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$=−\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}}\:\:+\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}}\:\:\:+\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:+\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$=−{H}_{{n}−\mathrm{1}} \:+{H}_{{n}+\mathrm{1}} \:−\frac{\mathrm{3}}{\mathrm{2}}\:+\:\:\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$={H}_{{n}+\mathrm{1}} −{H}_{{n}−\mathrm{1}} \:\:−\frac{\mathrm{11}}{\mathrm{4}}\:\:+\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:{but} \\ $$$${H}_{{n}+\mathrm{1}\:} \:−{H}_{{n}−\mathrm{1}} \:_{{n}\rightarrow\infty} \rightarrow\mathrm{0}\:\:\:\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)\:\rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\rightarrow\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow\:\mathrm{4}{S}_{{n}} \:\rightarrow\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:−\frac{\mathrm{11}}{\mathrm{4}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\:−\frac{\mathrm{11}}{\mathrm{16}}\:.\:{let}\:{remember}\:{that}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$$${and}\:\xi_{{n}} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} }\:. \\ $$$$ \\ $$

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