Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 45594 by maxmathsup by imad last updated on 14/Oct/18

calculate Σ_(n=1) ^∞ (1/((4n^2 −1)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by maxmathsup by imad last updated on 18/Oct/18

let S_n =Σ_(k=1) ^n (1/((4k^2 −1)^2 )) ⇒S_n =Σ_(k=1) ^n (1/((2k−1)^2 (2k+1)^2 )) let decompose F(x)=(1/((2x−1)^2 (2x+1)^2 )) ⇒F(x)=(a/(2x−1)) +(b/((2x−1)^2 )) +(c/(2x+1)) +(d/((2x+1)^2 )) b =lim_(x→(1/2)) (2x−1)^2 F(x)=(1/4) d =lim_(x→−(1/2)) (2x+1)^2 F(x) =(1/4) ⇒F(x)=(a/(2x−1)) +(1/(4(2x−1)^2 )) +(c/((2x+1))) +(d/(4(2x+)^2 )) lim_(x→+∞) x F(x)=0 =(a/2) +(c/2) ⇒c=−a ⇒ F(x)=(a/(2x−1)) −(a/(2x+1)) + (1/(4(2x−1)^2 )) +(1/(4(2x+1)^2 )) F(0)=1 =−2a +(1/2) ⇒(1/2) =−2a ⇒a=−(1/4) ⇒ F(x)=(1/(4(2x+1)))−(1/(4(2x−1))) +(1/(4(2x−1)^2 )) +(1/(4(2x+1)^2 )) ⇒ S_n =Σ_(k=1) ^n F(k)=(1/4) Σ_(k=1) ^n (1/(2k+1)) −(1/4)Σ_(k=1) ^n (1/(2k−1)) +(1/4)Σ_(k=1) ^n (1/((2k−1)^2 )) +(1/4)Σ_(k=1) ^n (1/((2k+1)^2 )) (1/4)Σ_(k=1) ^n {(1/(2k+1))−(1/(2k−1))} =(1/4)Σ_(k=1) ^n v_k −v_(k−1) =(1/4)(v_n −v_0 ) (v_n =(1/(2k+1))) =(1/4)((1/(2n+1)) −1) →−(1/4)(n→+∞) we have(1/4) {Σ_(k=1) ^n (1/((2k−1)^2 )) +Σ_(k=1) ^n (1/((2k+1)^2 ))}→(1/4){Σ_(n=1) ^∞ (1/((2n−1)^2 )) +Σ_(n=1) ^∞ (1/((2n+1)^2 ))} but Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (1/((2n+1)^2 )) =(π^2 /8) −1 Σ_(n=1) ^∞ (1/((2n−1)^2 )) =_(n=k+1) Σ_(k=0) ^∞ (1/((2k+1)^2 )) =(π^2 /8) ⇒ lim_(n→+∞) S_n =−(1/4) +(1/4){(π^2 /8)−1 +(π^2 /8)}=(1/4){(π^2 /4) −1)−(1/4) =(π^2 /(16)) −(1/2) .

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com