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Question Number 38518 by math khazana by abdo last updated on 26/Jun/18
calculate Σ_(n=1) ^∞ (1/(n^2 (2n−1)^2 ))
$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 27/Jun/18
let put S_n =Σ_(k=1) ^n (1/(k^2 (2k−1)^2 )) and let decompose F(x)= (1/(x^2 (2x−1)^2 )) F(x)=(a/x) +(b/x^2 ) +(c/(2x−1)) +(d/((2x−1)^2 )) b=lim_(x→0) x^2 F(x)=1 d=lim_(x→(1/2)) (2x−1)^2 F(x)=4 ⇒ F(x)=(a/x) +(1/x^2 ) +(c/(2x−1)) +(4/((2x−1)^2 )) lim_(x→+∞) x F(x)=0=a +(c/2) ⇒2a+c=0 ⇒c=−2a F(x)=(a/x) −((2a)/(2x−1)) +(1/x^2 ) +(4/((2x−1)^2 )) F(1)=1=a −2a +1 +4 ⇒−a +4=0 ⇒a=4 F(x)= (4/x) −(8/(2x−1)) +(1/x^2 ) +(4/((2x−1)^2 )) S_n =Σ_(k=1) ^n F(k) = 4Σ_(k=1) ^n (1/k) −8Σ_(k=1) ^n (1/(2k−1)) +Σ_(k=1) ^n (1/k^2 ) +4 Σ_(k=1) ^n (1/((2k−1)^2 )) but Σ_(k=1) ^n (1/k) =H_n Σ_(k=1) ^n (1/(2k−1)) =1 +(1/3) +(1/5) +....+(1/(2n−1)) =1+(1/2) +(1/3) +(1/4) +.....+(1/(2n−1)) +(1/(2n)) −(1/2)H_n =H_(2n) −(1/2) H_n Σ_(k=1) ^n (1/k^2 ) =ξ_n (2) Σ_(k=1) ^n (1/((2k−1)^2 )) =Σ_(k=2) ^(n+1) (1/((2k+1)^2 )) =Σ_(k=0) ^(n+1) (1/((2k+1)^2 )) ⇒ S_n = 4 H_n −8H_(2n) +4 H_n +ξ_n (2)+4Σ_(k=0) ^(n+1) (1/((2k+1)^2 )) S_n =−8(H_(2n) −H_n ) +ξ_n (2) +4Σ_(k=0) ^(n+1) (1/((2k+1)^2 )) lim_(n→+∞) S_n =−8ln(2) +(π^2 /6) +4 .(π^2 /8) =−8ln(2) +(π^2 /6) +(π^2 /2) =((4π^2 )/6) −8ln(2) =((2π^2 )/3) −8ln(2) .
$${let}\:{put}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\:\:{and}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{{d}}{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} \:{F}\left({x}\right)=\mathrm{1} \\ $$$${d}={lim}_{{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} \:\:\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\mathrm{4}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{\mathrm{4}}{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {x}\:{F}\left({x}\right)=\mathrm{0}={a}\:+\frac{{c}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{a}+{c}=\mathrm{0}\:\Rightarrow{c}=−\mathrm{2}{a} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:−\frac{\mathrm{2}{a}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{1}={a}\:−\mathrm{2}{a}\:\:+\mathrm{1}\:+\mathrm{4}\:\Rightarrow−{a}\:+\mathrm{4}=\mathrm{0}\:\Rightarrow{a}=\mathrm{4} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{4}}{{x}}\:−\frac{\mathrm{8}}{\mathrm{2}{x}−\mathrm{1}}\:\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} {F}\left({k}\right)\:=\:\mathrm{4}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\mathrm{8}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}} \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:+\mathrm{4}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{5}}\:+….+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+…..+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} ={H}_{\mathrm{2}{n}} \:−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}} \left(\mathrm{2}\right) \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}_{{n}} =\:\mathrm{4}\:{H}_{{n}} \:\:−\mathrm{8}{H}_{\mathrm{2}{n}} \:+\mathrm{4}\:{H}_{{n}} \:\:+\xi_{{n}} \left(\mathrm{2}\right)+\mathrm{4}\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${S}_{{n}} =−\mathrm{8}\left({H}_{\mathrm{2}{n}} \:−{H}_{{n}} \right)\:+\xi_{{n}} \left(\mathrm{2}\right)\:+\mathrm{4}\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{n}\rightarrow+\infty} {S}_{{n}} =−\mathrm{8}{ln}\left(\mathrm{2}\right)\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\mathrm{4}\:.\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$=−\mathrm{8}{ln}\left(\mathrm{2}\right)\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:=\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{6}}\:−\mathrm{8}{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{8}{ln}\left(\mathrm{2}\right)\:. \\ $$

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