show-that-n-2-1-n-2-1-n-2-2-0-2999-

Question Number 86786 by M±th+et£s last updated on 31/Mar/20

show that Σ_(n=2) ^∞ (1/(n^2 (1−n^2 )^2 ))=0.2999

$${show}\:{that} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left(\mathrm{1}−{n}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}.\mathrm{2999} \\ $$

Commented by Prithwish Sen 1 last updated on 31/Mar/20

$$\mathrm{I}\:\mathrm{got}\:\mathrm{0}.\mathrm{0299} \\ $$

Commented by Prithwish Sen 1 last updated on 31/Mar/20

the series is (3/2)Σ_1 ^∞ (1/n^2 ) −((39)/(16)) = (𝛑^2 /4) −((39)/(16)) = 0.0299011 please check.

$$\mathrm{the}\:\mathrm{series}\:\mathrm{is} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\underset{\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:−\frac{\mathrm{39}}{\mathrm{16}}\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{4}}\:−\frac{\mathrm{39}}{\mathrm{16}}\:=\:\mathrm{0}.\mathrm{0299011} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$

Commented by mathmax by abdo last updated on 31/Mar/20

let decompose F(x) =(1/(x^2 (x^2 −1)^2 )) =(1/(x^2 (x−1)^2 (x+1)^2 )) F(x) =(a/x)+(b/x^2 ) +(c/(x−1)) +(d/((x−1)^2 )) +(e/(x+1)) +(f/((x+1)^2 )) F(−x)=F(x) ⇒−(a/x) +(b/x^2 )−(c/(x+1)) +(d/((x+1)^2 ))−(e/(x−1)) +(f/((x−1)^2 )) =F(x) ⇒ a=0 and e=−c and f =d ⇒ f(x)=(b/x^2 ) +(c/(x−1)) +(d/((x−1)^2 ))−(c/(x+1)) +(d/((x+1)^2 )) b =1 and d =(1/4) ⇒F(x)=(1/x^2 ) +(c/(x−1)) +(1/(4(x−1)^2 ))−(c/(x+1)) +(1/(4(x+1)^2 )) F(2) =(1/(36)) =(1/4) +c +(1/4)−(c/3) +(1/(36)) ⇒(1/2) +(2/3)c =0 ⇒3+4c =0 ⇒ c=−(3/4) ⇒F(x) =(1/x^2 )−(3/(4(x−1))) +(1/(4(x−1)^2 )) +(3/(4(x+1)))+(1/(4(x+1)^2 )) S_n =Σ_(k=2) ^n (1/(k^2 (k^2 −1)^2 )) ⇒ S_n =Σ_(k=2) ^n (1/k^2 )−(3/4)Σ_(k=2) ^n (1/(k−1)) +(1/4)Σ_(k=2) ^n (1/((k−1)^2 )) +(3/4)Σ_(k=2) ^n (1/(k+1)) +(1/4)Σ_(k=2) ^n (1/((k+1)^2 )) Σ_(k.=2) ^n (1/k^2 ) =ξ_n (2)−1 →(π^2 /6)−1 Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) =H_(n−1) =ln(n−1)+γ +o((1/(n−1))) Σ_(k=2) ^n (1/((k−1)^2 )) =Σ_(k=1) ^(n−1) (1/k^2 ) =ξ_(n−1) (2) →(π^2 /6) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k) =H_(n+1) −(3/2) =ln(n+1)+γ +o((1/(n+1)))−(3/2) Σ_(k=2) ^n (1/((k+1)^2 )) =Σ_(k=3) ^(n+1) (1/k^2 ) =ξ_(n+1) (2)−(5/4)→(π^2 /6)−(5/4) ⇒ S_n =ξ_n (2)−1−(3/4)H_(n−1) +(1/4)ξ_(n−1) (2)+(3/4) H_(n+1) −(9/8) +(1/4)ξ_(n+1) (2)−(5/(16)) ⇒lim_(n→+∞) S_n =(π^2 /6)−1 +(1/4)(π^2 /6)−(9/8)+(1/4)(π^2 /6)−(5/(16)) =(π^2 /6)+(π^2 /(12)) −((16+18+5)/(16)) =(π^2 /4)−((39)/(16))

$${let}\:{decompose}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}−\mathrm{1}}\:+\frac{{d}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{e}}{{x}+\mathrm{1}}\:+\frac{{f}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(−{x}\right)={F}\left({x}\right)\:\Rightarrow−\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }−\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{{e}}{{x}−\mathrm{1}}\:+\frac{{f}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={F}\left({x}\right)\:\Rightarrow\:{a}=\mathrm{0}\:{and}\:{e}=−{c}\:\:{and}\:{f}\:={d}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}−\mathrm{1}}\:+\frac{{d}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}\:=\mathrm{1}\:\:{and}\:{d}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{{c}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{36}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:+{c}\:+\frac{\mathrm{1}}{\mathrm{4}}−\frac{{c}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{36}}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{2}}{\mathrm{3}}{c}\:=\mathrm{0}\:\Rightarrow\mathrm{3}+\mathrm{4}{c}\:=\mathrm{0}\:\Rightarrow \\ $$$${c}=−\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{4}}\sum_{{k}=\mathrm{2}} ^{{n}} \frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{k}.=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}} \left(\mathrm{2}\right)−\mathrm{1}\:\rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}−\mathrm{1}} ={ln}\left({n}−\mathrm{1}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}\right) \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)\:\rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{2}}\:={ln}\left({n}+\mathrm{1}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow \\ $$$${S}_{{n}} \:=\xi_{{n}} \left(\mathrm{2}\right)−\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}{H}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{4}}\:{H}_{{n}+\mathrm{1}} −\frac{\mathrm{9}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{4}}\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{16}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{9}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{16}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\frac{\mathrm{16}+\mathrm{18}+\mathrm{5}}{\mathrm{16}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{39}}{\mathrm{16}} \\ $$

Commented by M±th+et£s last updated on 31/Mar/20