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Question Number 36819 by maxmathsup by imad last updated on 06/Jun/18

find the value of the sum Σ_(n=1) ^∞ (1/((2n−1)^2 (2n+1)^2 ))

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

Commented by maxmathsup by imad last updated on 03/Aug/18

let S_n = Σ_(k=1) ^n (1/((2k−1)^2 (2k+1)^2 )) ⇒S=lim_(n→+∞) S_n 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))) +(1/(4(2x+1)^2 )) lim_(x→+∞) xF(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) ⇒2a =−(1/2) ⇒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)) −Σ_(k=1) ^n (1/(2k−1)) +Σ_(k=1) ^n (1/((2k−1)^2 )) +Σ_(k=1) ^n (1/((2k+1)^2 ))}but Σ_(k=1) ^n ((1/(2k+1)) −(1/(2k−1))) =Σ_(k=1) ^n (u_n −u_(n−1) ) (u_n =(1/(2n+1))) =u_n −u_0 =(1/(2n+1)) −1 also Σ_(k=1) ^n (1/((2k−1)^2 )) =_(k=j+1) Σ_(j=0) ^(n−1) (1/((2j+1)^2 )) →Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(k=1) ^n (1/((2k+1)^2 )) =Σ_(k=0) ^n (1/((2k+1)^2 )) −1 →Σ_(n=0) ^∞ (1/((2n+1)^2 )) −1 we have (π^2 /6) =Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/(4n^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4) (π^2 /6) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /6) −(π^2 /(24)) =((3π^2 )/(24)) =(π^2 /8) ⇒ lim_ S_n =(1/4){−1 +(π^2 /8) +(π^2 /8) −1} =(1/4){(π^2 /4) −2} =(π^2 /(16)) −(1/2) S=(π^2 /(16)) −(1/2) .

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

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