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Question Number 36820 by maxmathsup by imad last updated on 06/Jun/18
find the value of Σ_(n=2) ^∞ (1/((n−1)^2 (n+1)^2 ))
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$
Commented by math khazana by abdo last updated on 24/Jun/18
let S_n =Σ_(k=2S) ^n (1/((n−1)^2 (n+1)^2 )) we have S=lim_(n→+∞) S_n let decompose F(x)= (1/((x−1)^2 (x+1)^2 )) F(x)= (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)) −(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) ⇒ F(x)= (1/(4(x+1))) −(1/(4(x−1))) +(1/(4(x−1)^2 )) +(1/(4(x+1)^2 )) 4S_n = 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 )) but Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k)= H_(n+1) −(3/2) Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) =H_(n−1) Σ_(k=2) ^n (1/((k−1)^2 )) = Σ_(k=1) ^(n−1) (1/k^2 ) =ξ_(n−1) (2) Σ_(k=2) ^n (1/((k+1)^2 )) =Σ_(k=3) ^(n+1) (1/k^2 ) =ξ_(n+1) (2)−(5/4) ⇒ 4S_n = H_(n+1) −(3/2) −H_(n−1) +ξ_(n−1) (2) +ξ_(n+1) (2)−(5/4) 4S_n =γ +ln(n+1) +o((1/n))−γ−ln(n−1)−o((1/n)) +ξ_(n−1) (2)+ξ_(n+1) (2) −((11)/4) ⇒ 4 S_n → ((2π^2 )/6) −((11)/4) =(π^2 /3) −((11)/4) ⇒ lim_(n→+∞) S_n = (π^2 /(12)) −((11)/(16)) = S.
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{2}{S}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have}\:{S}={lim}_{{n}\rightarrow+\infty} {S}_{{n}} \\ $$$${let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${d}={lim}_{{x}\rightarrow−\mathrm{1}} \:\left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:−\frac{{a}}{{x}+\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}=−\mathrm{2}{a}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow−\mathrm{2}{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}{S}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:+\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{k}}=\:{H}_{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}−\mathrm{1}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{4}{S}_{{n}} =\:{H}_{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{2}}\:−{H}_{{n}−\mathrm{1}} \:\:+\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{4}{S}_{{n}} =\gamma\:+{ln}\left({n}+\mathrm{1}\right)\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)−\gamma−{ln}\left({n}−\mathrm{1}\right)−{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$+\xi_{{n}−\mathrm{1}} \left(\mathrm{2}\right)+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:−\frac{\mathrm{11}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{4}\:{S}_{{n}} \:\rightarrow\:\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{11}}{\mathrm{4}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:−\frac{\mathrm{11}}{\mathrm{4}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\frac{\mathrm{11}}{\mathrm{16}}\:=\:{S}. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18
T_n =(1/((n+1)^2 (n−1)^2 ))=(1/2)(((n+1)−(n−1))/((n+1)^2 (n−1)^2 )) =(1/2)(1/((n+1)(n−1)^2 ))−(1/2)(1/((n+1)^2 (n−1))) =(1/4)×(((n+1)−(n−1))/((n+1)(n−1)^2 ))−(1/4)×(((n+1)−(n−1))/((n+1)^2 (n−1))) =(1/4)×(1/((n−1)^2 ))−(1/4)×(1/((n+1)(n−1)))− (1/4)×(1/((n+1)(n−1)))+(1/4)×(1/((n+1)^2 )) =(1/4)×(1/((n−1)^2 ))−(1/2)×(1/((n+1)(n−1)))+(1/4)×(1/((n+1)^2 )) so S_n =S_1 −S_2 +S_3 (n>1) S_1 =(1/4)((1/1^2 )+(1/2^2 )+(1/3^2 )+.....)=(1/4)×(Π^2 /6) S_2 =(1/4){(((n+1)−(n−1))/((n+1)(n−1)))} =(1/4){(1/(n−1))−(1/(n+1))} ={(1/4)((1/1)+(1/2)+(1/3)+...)−(1/4)((1/3)+(1/4)+(1/5)+(1/6)...)} =(1/4)×(3/2)=(3/8) S_3 =(1/4)×{(1/((n+1)^2 ))} =(1/4)×((1/3^2 )+(1/4^2 )+(1/5^2 )+....) =(1/4)×((Π^2 /6)−(1/1^2 )−(1/2^2 ))=(1/4)((Π^2 /6)−(5/4)) S_n =S_1 −S_2 +S_3 =(1/4)×(Π^2 /6)−(1/4)×(3/2)+(1/4)((Π^2 /6)−(5/4)) =(1/4)×{(Π^2 /6)−(3/2)+(Π^2 /6)−(5/4)} =(1/4)×{((2Π^2 )/6)−((11)/4)}
$${T}_{{n}} =\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}− \\ $$$$\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${so} \\ $$$${S}_{{n}} ={S}_{\mathrm{1}} −{S}_{\mathrm{2}} +{S}_{\mathrm{3}} \:\:\:\:\:\:\left({n}>\mathrm{1}\right) \\ $$$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…..\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\prod^{\mathrm{2}} }{\mathrm{6}} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\} \\ $$$$=\left\{\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}…\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${S}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{4}}×\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+….\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\left(\frac{\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$${S}_{{n}} ={S}_{\mathrm{1}} −{S}_{\mathrm{2}} +{S}_{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\left\{\frac{\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\left\{\frac{\mathrm{2}\prod^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{11}}{\mathrm{4}}\right\} \\ $$
Commented by math khazana by abdo last updated on 24/Jun/18
correct answer sir Tanmay...
$${correct}\:{answer}\:{sir}\:{Tanmay}… \\ $$

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